Class 12: Indefinite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int \frac{2x+1}{(x+1)(x-2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{2x+1}{(x+1)(x-2)}\,dx$
$\displaystyle \text{Let } \frac{2x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\qquad (1)$
$\displaystyle \Rightarrow \frac{2x+1}{(x+1)(x-2)}=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$
$\displaystyle \text{Then, } 2x+1=A(x-2)+B(x+1)\qquad (2)$
$\displaystyle \text{Putting } x=2 \text{ in eq. (2)}$
$\displaystyle 2(2)+1=A(0)+B(3)$
$\displaystyle \Rightarrow B=\frac{5}{3}$
$\displaystyle \text{Putting } x=-1 \text{ in eq. (2)}$
$\displaystyle 2(-1)+1=A(-3)+B(0)$
$\displaystyle \Rightarrow -1=-3A$
$\displaystyle \Rightarrow A=\frac{1}{3}$
$\displaystyle \text{Substituting the values of } A \text{ and } B \text{ in eq. (1), we get}$
$\displaystyle \frac{2x+1}{(x+1)(x-2)}=\frac{1}{3}\cdot\frac{1}{x+1}+\frac{5}{3}\cdot\frac{1}{x-2}$
$\displaystyle \int \frac{2x+1}{(x+1)(x-2)}\,dx=\frac{1}{3}\int\frac{1}{x+1}\,dx+\frac{5}{3}\int\frac{1}{x-2}\,dx$
$\displaystyle =\frac{1}{3}\log|x+1|+\frac{5}{3}\log|x-2|+C$

$\displaystyle \textbf{Question 2: }~\int \frac{1}{x(x-2)(x-4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{1}{x(x-2)(x-4)}\,dx$
$\displaystyle \text{Let } \frac{1}{x(x-2)(x-4)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4}$
$\displaystyle \Rightarrow \frac{1}{x(x-2)(x-4)}=\frac{A(x-2)(x-4)+Bx(x-4)+Cx(x-2)}{x(x-2)(x-4)}$
$\displaystyle \Rightarrow 1=A(x-2)(x-4)+Bx(x-4)+Cx(x-2)\qquad (1)$
$\displaystyle \text{Putting } x=0 \text{ in eq. (1)}$
$\displaystyle 1=A(-2)(-4)+B(0)+C(0)$
$\displaystyle \Rightarrow A=\frac{1}{8}$
$\displaystyle \text{Putting } x=2 \text{ in eq. (1)}$
$\displaystyle 1=A(0)+B(2)(-2)+C(0)$
$\displaystyle \Rightarrow B=-\frac{1}{4}$
$\displaystyle \text{Putting } x=4 \text{ in eq. (1)}$
$\displaystyle 1=A(0)+B(0)+C(4)(2)$
$\displaystyle \Rightarrow C=\frac{1}{8}$
$\displaystyle \Rightarrow \frac{1}{x(x-2)(x-4)}=\frac{1}{8x}-\frac{1}{4(x-2)}+\frac{1}{8(x-4)}$
$\displaystyle \int \frac{1}{x(x-2)(x-4)}\,dx=\frac{1}{8}\int\frac{1}{x}\,dx-\frac{1}{4}\int\frac{1}{x-2}\,dx+\frac{1}{8}\int\frac{1}{x-4}\,dx$
$\displaystyle =\frac{1}{8}\log|x|-\frac{1}{4}\log|x-2|+\frac{1}{8}\log|x-4|+C$
$\displaystyle =\frac{1}{8}\left(\log|x|+\log|x-4|-2\log|x-2|\right)+C$
$\displaystyle =\frac{1}{8}\log\left|\frac{x(x-4)}{(x-2)^2}\right|+C$

$\displaystyle \textbf{Question 3: }~\int \frac{x^2+x-1}{x^2+x-6}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x^2+x-1}{x^2+x-6}\,dx$
$\displaystyle =\int \frac{x^2+x-6+5}{x^2+x-6}\,dx$
$\displaystyle =\int \frac{x^2+x-6}{x^2+x-6}\,dx+5\int \frac{1}{x^2+x-6}\,dx$
$\displaystyle =\int dx+5\int \frac{1}{x^2+x-6}\,dx$
$\displaystyle =\int dx+5\int \frac{1}{(x-2)(x+3)}\,dx\qquad (1)$
$\displaystyle \text{Let } \frac{1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}$
$\displaystyle \Rightarrow \frac{1}{(x-2)(x+3)}=\frac{A(x+3)+B(x-2)}{(x-2)(x+3)}$
$\displaystyle \Rightarrow 1=A(x+3)+B(x-2)\qquad (2)$
$\displaystyle \text{Putting } x=-3 \text{ in eq. (2)}$
$\displaystyle 1=A(0)+B(-5)$
$\displaystyle \Rightarrow B=-\frac{1}{5}$
$\displaystyle \text{Putting } x=2 \text{ in eq. (2)}$
$\displaystyle 1=A(5)+B(0)$
$\displaystyle \Rightarrow A=\frac{1}{5}$
$\displaystyle \Rightarrow \frac{1}{(x-2)(x+3)}=\frac{1}{5}\cdot\frac{1}{x-2}-\frac{1}{5}\cdot\frac{1}{x+3}$
$\displaystyle \int \frac{1}{(x-2)(x+3)}\,dx=\frac{1}{5}\int\frac{1}{x-2}\,dx-\frac{1}{5}\int\frac{1}{x+3}\,dx$
$\displaystyle =\frac{1}{5}\log|x-2|-\frac{1}{5}\log|x+3|+C$
$\displaystyle =\frac{1}{5}\log\left|\frac{x-2}{x+3}\right|+C\qquad (3)$
$\displaystyle \text{From eq. (1) and eq. (3)}$
$\displaystyle \int \frac{x^2+x-1}{x^2+x-6}\,dx=x+\log|x-2|-\log|x+3|+C$

$\displaystyle \textbf{Question 4: }~\int \frac{3+4x-x^2}{(x+2)(x-1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{3+4x-x^2}{x^2-x+2x-2}\,dx$
$\displaystyle =\int \frac{-x^2+4x+3}{x^2+x-2}\,dx$
$\displaystyle \frac{-x^2+4x+3}{x^2+x-2}=-1+\frac{5x+1}{x^2+x-2}\qquad (1)$
$\displaystyle \text{Now } \frac{5x+1}{x^2+x-2}=\frac{5x+1}{(x-1)(x+2)}$
$\displaystyle \text{Let } \frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$
$\displaystyle \Rightarrow \frac{5x+1}{(x-1)(x+2)}=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$
$\displaystyle \Rightarrow 5x+1=A(x+2)+B(x-1)\qquad (2)$
$\displaystyle \text{Putting } x=-2 \text{ in eq. (2)}$
$\displaystyle 5(-2)+1=A(0)+B(-3)$
$\displaystyle \Rightarrow B=3$
$\displaystyle \text{Putting } x=1 \text{ in eq. (2)}$
$\displaystyle 5(1)+1=A(3)+B(0)$
$\displaystyle \Rightarrow A=2$
$\displaystyle \Rightarrow \frac{5x+1}{(x-1)(x+2)}=\frac{2}{x-1}+\frac{3}{x+2}\qquad (3)$
$\displaystyle \text{From eq. (1) and eq. (3)}$
$\displaystyle \frac{-x^2+4x+3}{x^2+x-2}=-1+\frac{2}{x-1}+\frac{3}{x+2}$
$\displaystyle \int \frac{-x^2+4x+3}{x^2+x-2}\,dx=\int(-1)\,dx+\int\frac{2}{x-1}\,dx+\int\frac{3}{x+2}\,dx$
$\displaystyle =-x+2\log|x-1|+3\log|x+2|+C$

$\displaystyle \textbf{Question 5: }~\int \frac{x^2+1}{x^2-1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x^2+1}{x^2-1}\,dx$
$\displaystyle =\int \frac{x^2-1+2}{x^2-1}\,dx$
$\displaystyle =\int dx+2\int \frac{1}{x^2-1}\,dx$
$\displaystyle =\int dx+2\int \frac{1}{(x-1)(x+1)}\,dx\qquad (1)$
$\displaystyle \text{Let } \frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$
$\displaystyle \Rightarrow \frac{1}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}$
$\displaystyle \Rightarrow 1=A(x+1)+B(x-1)\qquad (2)$
$\displaystyle \text{Putting } x=-1 \text{ in eq. (2)}$
$\displaystyle 1=A(0)+B(-2)$
$\displaystyle \Rightarrow B=-\frac{1}{2}$
$\displaystyle \text{Putting } x=1 \text{ in eq. (2)}$
$\displaystyle 1=A(2)+B(0)$
$\displaystyle \Rightarrow A=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1}{(x-1)(x+1)}=\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\qquad (3)$
$\displaystyle \text{From eq. (1) and eq. (3)}$
$\displaystyle \int \frac{x^2+1}{x^2-1}\,dx=\int dx+2\int\left[\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\right]dx$
$\displaystyle =\int dx+\int\frac{1}{x-1}\,dx-\int\frac{1}{x+1}\,dx$
$\displaystyle =x+\log|x-1|-\log|x+1|+C$
$\displaystyle =x+\log\left|\frac{x-1}{x+1}\right|+C$

$\displaystyle \textbf{Question 6: }~\int \frac{x^2}{(x-1)(x-2)(x-3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x^2}{(x-1)(x-2)(x-3)}\,dx$
$\displaystyle \text{Let } \frac{x^2}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
$\displaystyle \Rightarrow \frac{x^2}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$
$\displaystyle \Rightarrow x^2=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\qquad (1)$
$\displaystyle \text{Putting } x=1 \text{ in eq. (1)}$
$\displaystyle 1=A(-1)(-2)$
$\displaystyle \Rightarrow A=\frac{1}{2}$
$\displaystyle \text{Putting } x=2 \text{ in eq. (1)}$
$\displaystyle 4=B(1)(-1)$
$\displaystyle \Rightarrow B=-4$
$\displaystyle \text{Putting } x=3 \text{ in eq. (1)}$
$\displaystyle 9=C(2)(1)$
$\displaystyle \Rightarrow C=\frac{9}{2}$
$\displaystyle \Rightarrow \frac{x^2}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{4}{x-2}+\frac{9}{2(x-3)}$
$\displaystyle \int \frac{x^2}{(x-1)(x-2)(x-3)}\,dx=\frac{1}{2}\int\frac{1}{x-1}\,dx-4\int\frac{1}{x-2}\,dx+\frac{9}{2}\int\frac{1}{x-3}\,dx$
$\displaystyle =\frac{1}{2}\log|x-1|-4\log|x-2|+\frac{9}{2}\log|x-3|+C$

$\displaystyle \textbf{Question 7: }~\int \frac{5x}{(x+1)(x^2-4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{5x}{(x+1)(x^2-4)}\,dx$
$\displaystyle =\int \frac{5x}{(x+1)(x-2)(x+2)}\,dx$
$\displaystyle \text{Let } \frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}$
$\displaystyle \Rightarrow \frac{5x}{(x+1)(x-2)(x+2)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$
$\displaystyle \Rightarrow 5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)\qquad (1)$
$\displaystyle \text{Putting } x=2 \text{ in eq. (1)}$
$\displaystyle 5(2)=B(3)(4)$
$\displaystyle \Rightarrow B=\frac{5}{6}$
$\displaystyle \text{Putting } x=-2 \text{ in eq. (1)}$
$\displaystyle 5(-2)=C(-1)(-4)$
$\displaystyle \Rightarrow C=-\frac{5}{2}$
$\displaystyle \text{Putting } x=-1 \text{ in eq. (1)}$
$\displaystyle 5(-1)=A(-3)(1)$
$\displaystyle \Rightarrow A=\frac{5}{3}$
$\displaystyle \Rightarrow \frac{5x}{(x+1)(x-2)(x+2)}=\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}$
$\displaystyle \int \frac{5x}{(x+1)(x-2)(x+2)}\,dx=\frac{5}{3}\int\frac{1}{x+1}\,dx+\frac{5}{6}\int\frac{1}{x-2}\,dx-\frac{5}{2}\int\frac{1}{x+2}\,dx$
$\displaystyle =\frac{5}{3}\log|x+1|+\frac{5}{6}\log|x-2|-\frac{5}{2}\log|x+2|+C$
$\displaystyle =\frac{5}{6}\left(2\log|x+1|+\log|x-2|-3\log|x+2|\right)+C$
$\displaystyle =\frac{5}{6}\log\left|\frac{(x+1)^2(x-2)}{(x+2)^3}\right|+C$

$\displaystyle \textbf{Question 8: }~\int \frac{x^2+1}{x(x^2-1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x^2+1}{x(x^2-1)}\,dx$
$\displaystyle =\int \frac{x^2+1}{x(x-1)(x+1)}\,dx$
$\displaystyle \text{Let } \frac{x^2+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$
$\displaystyle \Rightarrow \frac{x^2+1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+Bx(x+1)+Cx(x-1)}{x(x-1)(x+1)}$
$\displaystyle \Rightarrow x^2+1=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)\qquad (1)$
$\displaystyle \text{Putting } x=1 \text{ in eq. (1)}$
$\displaystyle 1^2+1=A(0)+B(1)(2)+C(0)$
$\displaystyle \Rightarrow B=1$
$\displaystyle \text{Putting } x=0 \text{ in eq. (1)}$
$\displaystyle 0^2+1=A(-1)(1)+B(0)+C(0)$
$\displaystyle \Rightarrow A=-1$
$\displaystyle \text{Putting } x=-1 \text{ in eq. (1)}$
$\displaystyle (-1)^2+1=A(0)+B(0)+C(-1)(-2)$
$\displaystyle \Rightarrow C=1$
$\displaystyle \Rightarrow \frac{x^2+1}{x(x-1)(x+1)}=-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x+1}$
$\displaystyle \int \frac{x^2+1}{x(x^2-1)}\,dx=-\int\frac{1}{x}\,dx+\int\frac{1}{x-1}\,dx+\int\frac{1}{x+1}\,dx$
$\displaystyle =-\log|x|+\log|x-1|+\log|x+1|+C$
$\displaystyle =\log\left|\frac{x^2-1}{x}\right|+C$

$\displaystyle \textbf{Question 9: }~\int \frac{2x-3}{(x^2-1)(2x+3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{2x-3}{(x^2-1)(2x+3)}\,dx$
$\displaystyle =\int \frac{2x-3}{(x-1)(x+1)(2x+3)}\,dx$
$\displaystyle \text{Let } \frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}$
$\displaystyle \Rightarrow \frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2x+3)}$
$\displaystyle \Rightarrow 2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)\qquad (1)$
$\displaystyle \text{Putting } x=-1 \text{ in eq. (1)}$
$\displaystyle 2(-1)-3=B(-2)(1)$
$\displaystyle \Rightarrow -5=-2B$
$\displaystyle \Rightarrow B=\frac{5}{2}$
$\displaystyle \text{Putting } x=1 \text{ in eq. (1)}$
$\displaystyle 2(1)-3=A(2)(5)$
$\displaystyle \Rightarrow -1=10A$
$\displaystyle \Rightarrow A=-\frac{1}{10}$
$\displaystyle \text{Putting } 2x+3=0 \text{ or } x=-\frac{3}{2} \text{ in eq. (1)}$
$\displaystyle 2\left(-\frac{3}{2}\right)-3=C\left(-\frac{5}{2}\right)\left(-\frac{1}{2}\right)$
$\displaystyle \Rightarrow -6=\frac{5}{4}C$
$\displaystyle \Rightarrow C=-\frac{24}{5}$
$\displaystyle \Rightarrow \frac{2x-3}{(x-1)(x+1)(2x+3)}=-\frac{1}{10(x-1)}+\frac{5}{2(x+1)}-\frac{24}{5(2x+3)}$
$\displaystyle \int \frac{2x-3}{(x-1)(x+1)(2x+3)}\,dx=-\frac{1}{10}\int\frac{1}{x-1}\,dx+\frac{5}{2}\int\frac{1}{x+1}\,dx-\frac{24}{5}\int\frac{1}{2x+3}\,dx$
$\displaystyle =-\frac{1}{10}\log|x-1|+\frac{5}{2}\log|x+1|-\frac{12}{5}\log|2x+3|+C$

$\displaystyle \textbf{Question 10: }~\int \frac{x^3}{(x-1)(x-2)(x-3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x^3}{(x-1)(x-2)(x-3)}\,dx$
$\displaystyle =\int \frac{x^3}{(x-1)(x^2-5x+6)}\,dx$
$\displaystyle =\int \frac{x^3}{x^3-6x^2+11x-6}\,dx$
$\displaystyle \frac{x^3}{x^3-6x^2+11x-6}=1+\frac{6x^2-11x+6}{x^3-6x^2+11x-6}$
$\displaystyle \Rightarrow \int \frac{x^3}{(x-1)(x-2)(x-3)}\,dx=\int dx+\int \frac{6x^2-11x+6}{(x-1)(x-2)(x-3)}\,dx\qquad (1)$
$\displaystyle \text{Let } \frac{6x^2-11x+6}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
$\displaystyle \Rightarrow \frac{6x^2-11x+6}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$
$\displaystyle \Rightarrow 6x^2-11x+6=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\qquad (2)$
$\displaystyle \text{Putting } x=2 \text{ in eq. (2)}$
$\displaystyle 6(2)^2-11(2)+6=B(1)(-1)$
$\displaystyle \Rightarrow 8=-B$
$\displaystyle \Rightarrow B=-8$
$\displaystyle \text{Putting } x=3 \text{ in eq. (2)}$
$\displaystyle 6(3)^2-11(3)+6=C(2)(1)$
$\displaystyle \Rightarrow 27=2C$
$\displaystyle \Rightarrow C=\frac{27}{2}$
$\displaystyle \text{Putting } x=1 \text{ in eq. (2)}$
$\displaystyle 6(1)^2-11(1)+6=A(-1)(-2)$
$\displaystyle \Rightarrow 1=2A$
$\displaystyle \Rightarrow A=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{6x^2-11x+6}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{8}{x-2}+\frac{27}{2(x-3)}\qquad (3)$
$\displaystyle \text{From eq. (1) and eq. (3)}$
$\displaystyle \int \frac{x^3}{(x-1)(x-2)(x-3)}\,dx=\int dx+\frac{1}{2}\int\frac{1}{x-1}\,dx-8\int\frac{1}{x-2}\,dx+\frac{27}{2}\int\frac{1}{x-3}\,dx$
$\displaystyle =x+\frac{1}{2}\log|x-1|-8\log|x-2|+\frac{27}{2}\log|x-3|+C$

$\displaystyle \textbf{Question 11: }~\int \frac{\sin 2x}{(1+\sin x)(2+\sin x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{\sin 2x\,dx}{(1+\sin x)(2+\sin x)}$
$\displaystyle =\int \frac{2\sin x\cos x\,dx}{(1+\sin x)(2+\sin x)}$
$\displaystyle \text{Put } t=\sin x$
$\displaystyle \Rightarrow dt=\cos x\,dx$
$\displaystyle \Rightarrow I=\int \frac{2t\,dt}{(1+t)(2+t)}$
$\displaystyle =2\int \frac{t}{(1+t)(2+t)}\,dt$
$\displaystyle \text{Let } \frac{t}{(1+t)(2+t)}=\frac{A}{t+1}+\frac{B}{t+2}$
$\displaystyle \Rightarrow \frac{t}{(1+t)(2+t)}=\frac{A(t+2)+B(t+1)}{(1+t)(2+t)}$
$\displaystyle \Rightarrow t=A(t+2)+B(t+1)$
$\displaystyle \text{Putting } t=-2$
$\displaystyle -2=B(-1)$
$\displaystyle \Rightarrow B=2$
$\displaystyle \text{Putting } t=-1$
$\displaystyle -1=A(1)$
$\displaystyle \Rightarrow A=-1$
$\displaystyle \Rightarrow I=2\int\left(-\frac{1}{t+1}+\frac{2}{t+2}\right)dt$
$\displaystyle =-2\log|t+1|+4\log|t+2|+C$
$\displaystyle =\log\left|\frac{(t+2)^4}{(t+1)^2}\right|+C$
$\displaystyle =\log\left|\frac{(\sin x+2)^4}{(\sin x+1)^2}\right|+C$

$\displaystyle \textbf{Question 12: }~\int \frac{2x}{(x^2+1)(x^2+3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{2x\,dx}{(x^2+1)(x^2+3)}$
$\displaystyle \text{Put } t=x^2$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow I=\int \frac{dt}{(t+1)(t+3)}$
$\displaystyle \text{Let } \frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}$
$\displaystyle \Rightarrow \frac{1}{(t+1)(t+3)}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}$
$\displaystyle \Rightarrow 1=A(t+3)+B(t+1)$
$\displaystyle \text{Putting } t=-3$
$\displaystyle 1=B(-2)$
$\displaystyle \Rightarrow B=-\frac{1}{2}$
$\displaystyle \text{Putting } t=-1$
$\displaystyle 1=A(2)$
$\displaystyle \Rightarrow A=\frac{1}{2}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\frac{dt}{t+1}-\frac{1}{2}\int\frac{dt}{t+3}$
$\displaystyle =\frac{1}{2}\log|t+1|-\frac{1}{2}\log|t+3|+C$
$\displaystyle =\frac{1}{2}\log\left|\frac{t+1}{t+3}\right|+C$
$\displaystyle =\frac{1}{2}\log\left|\frac{x^2+1}{x^2+3}\right|+C$

$\displaystyle \textbf{Question 13: }~\int \frac{1}{x\log x(2+\log x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{x\log x\,(2+\log x)}$
$\displaystyle \text{Put } t=\log x$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt$
$\displaystyle \Rightarrow I=\int \frac{dt}{t(t+2)}$
$\displaystyle \text{Let } \frac{1}{t(t+2)}=\frac{A}{t}+\frac{B}{t+2}$
$\displaystyle \Rightarrow \frac{1}{t(t+2)}=\frac{A(t+2)+Bt}{t(t+2)}$
$\displaystyle \Rightarrow 1=A(t+2)+Bt$
$\displaystyle \text{Putting } t=-2$
$\displaystyle 1=B(-2)$
$\displaystyle \Rightarrow B=-\frac{1}{2}$
$\displaystyle \text{Putting } t=0$
$\displaystyle 1=A(2)$
$\displaystyle \Rightarrow A=\frac{1}{2}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\frac{dt}{t}-\frac{1}{2}\int\frac{dt}{t+2}$
$\displaystyle =\frac{1}{2}\log|t|-\frac{1}{2}\log|t+2|+C$
$\displaystyle =\frac{1}{2}\log\left|\frac{t}{t+2}\right|+C$
$\displaystyle =\frac{1}{2}\log\left|\frac{\log x}{\log x+2}\right|+C$

$\displaystyle \textbf{Question 14: }~\int \frac{x^2+x+1}{(x^2+1)(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x^2+x+1}{(x^2+1)(x+2)}\,dx$
$\displaystyle \text{We express } \frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow x^2+x+1=A(x^2+1)+(Bx+C)(x+2)$
$\displaystyle \Rightarrow x^2+x+1=A x^2+A+Bx^2+2Bx+Cx+2C$
$\displaystyle \Rightarrow x^2+x+1=(A+B)x^2+(2B+C)x+(A+2C)$
$\displaystyle \text{Equating coefficients of } x^2,x \text{ and constants}$
$\displaystyle A+B=1$
$\displaystyle 2B+C=1$
$\displaystyle A+2C=1$
$\displaystyle \Rightarrow A=\frac{3}{5},\; B=\frac{2}{5},\; C=\frac{1}{5}$
$\displaystyle \Rightarrow I=\int\left(\frac{3}{5}\cdot\frac{1}{x+2}+\frac{2}{5}\cdot\frac{x}{x^2+1}+\frac{1}{5}\cdot\frac{1}{x^2+1}\right)dx$
$\displaystyle =\frac{3}{5}\int\frac{1}{x+2}\,dx+\frac{2}{5}\int\frac{x}{x^2+1}\,dx+\frac{1}{5}\int\frac{1}{x^2+1}\,dx\qquad (1)$
$\displaystyle \text{Now, } I_1=\int\frac{1}{x+2}\,dx$
$\displaystyle \text{Put } u=x+2$
$\displaystyle \Rightarrow du=dx$
$\displaystyle \Rightarrow I_1=\int\frac{1}{u}\,du$
$\displaystyle =\log|u|+c_1$
$\displaystyle =\log|x+2|+c_1\qquad (2)$
$\displaystyle \text{Now, } I_2=\int\frac{x}{x^2+1}\,dx$
$\displaystyle \text{Put } u=x^2+1$
$\displaystyle \Rightarrow du=2x\,dx$
$\displaystyle \Rightarrow I_2=\frac{1}{2}\int\frac{1}{u}\,du$
$\displaystyle =\frac{1}{2}\log|u|+c_2$
$\displaystyle =\frac{1}{2}\log|x^2+1|+c_2\qquad (3)$
$\displaystyle \text{Now, } I_3=\int\frac{1}{x^2+1}\,dx$
$\displaystyle =\tan^{-1}x+c_3\qquad (4)$
$\displaystyle \text{From (1), (2), (3) and (4)}$
$\displaystyle I=\frac{3}{5}\log|x+2|+\frac{2}{5}\cdot\frac{1}{2}\log|x^2+1|+\frac{1}{5}\tan^{-1}x+C$
$\displaystyle =\frac{3}{5}\log|x+2|+\frac{1}{5}\log|x^2+1|+\frac{1}{5}\tan^{-1}x+C$
$\displaystyle \therefore \int \frac{x^2+x+1}{(x^2+1)(x+2)}\,dx=\frac{3}{5}\log|x+2|+\frac{1}{5}\log|x^2+1|+\frac{1}{5}\tan^{-1}x+C$

$\displaystyle \textbf{Question 15: }~\int \frac{ax^2+bx+c}{(x-a)(x-b)(x-c)}\,dx,\ \text{where } a,b,c \text{ are distinct}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{ax^2+bx+c}{(x-a)(x-b)(x-c)}\,dx$
$\displaystyle \text{Let } \frac{ax^2+bx+c}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$
$\displaystyle \Rightarrow ax^2+bx+c=A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)$
$\displaystyle \Rightarrow ax^2+bx+c=A[x^2-(b+c)x+bc]+B[x^2-(c+a)x+ca]+C[x^2-(a+b)x+ab]$
$\displaystyle \Rightarrow ax^2+bx+c=(A+B+C)x^2-[A(b+c)+B(c+a)+C(a+b)]x+Abc+Bca+Cab$
$\displaystyle \text{Equating coefficients of } x^2,x \text{ and constants}$
$\displaystyle a=A+B+C\qquad (1)$
$\displaystyle b=-[A(b+c)+B(c+a)+C(a+b)]\qquad (2)$
$\displaystyle c=Abc+Bca+Cab\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{a^2a+ab+c}{(a-b)(a-c)}$
$\displaystyle B=\frac{ab^2+b^2+c}{(b-a)(b-c)}$
$\displaystyle C=\frac{ac^2+bc+c}{(c-a)(c-b)}$
$\displaystyle \Rightarrow I=\int\left(\frac{a^2a+ab+c}{(a-b)(a-c)}\cdot\frac{1}{x-a}+\frac{ab^2+b^2+c}{(b-a)(b-c)}\cdot\frac{1}{x-b}+\frac{ac^2+bc+c}{(c-a)(c-b)}\cdot\frac{1}{x-c}\right)dx$
$\displaystyle =\frac{a^2a+ab+c}{(a-b)(a-c)}\log|x-a|+\frac{ab^2+b^2+c}{(b-a)(b-c)}\log|x-b|+\frac{ac^2+bc+c}{(c-a)(c-b)}\log|x-c|+C$

$\displaystyle \textbf{Question 16: }~\int \frac{x}{(x^2+1)(x-1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x}{(x^2+1)(x-1)}\,dx$
$\displaystyle \text{We express } \frac{x}{(x^2+1)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow x=A(x^2+1)+(Bx+C)(x-1)$
$\displaystyle \Rightarrow x=A x^2+A+Bx^2-Bx+Cx-C$
$\displaystyle \Rightarrow x=(A+B)x^2+(-B+C)x+(A-C)$
$\displaystyle \text{Equating coefficients of } x^2,x \text{ and constants}$
$\displaystyle A+B=0$
$\displaystyle -B+C=1$
$\displaystyle A-C=0$
$\displaystyle \Rightarrow A=\frac{1}{2},\; B=-\frac{1}{2},\; C=\frac{1}{2}$
$\displaystyle \Rightarrow I=\int\left(\frac{1}{2}\cdot\frac{1}{x-1}-\frac{1}{2}\cdot\frac{x}{x^2+1}+\frac{1}{2}\cdot\frac{1}{x^2+1}\right)dx$
$\displaystyle =\frac{1}{2}\int\frac{1}{x-1}\,dx-\frac{1}{2}\int\frac{x}{x^2+1}\,dx+\frac{1}{2}\int\frac{1}{x^2+1}\,dx\qquad (1)$
$\displaystyle \text{Now, } I_1=\int\frac{1}{x-1}\,dx$
$\displaystyle \text{Put } u=x-1$
$\displaystyle \Rightarrow du=dx$
$\displaystyle \Rightarrow I_1=\int\frac{1}{u}\,du$
$\displaystyle =\log|u|+c_1$
$\displaystyle =\log|x-1|+c_1\qquad (2)$
$\displaystyle \text{Now, } I_2=\int\frac{x}{x^2+1}\,dx$
$\displaystyle \text{Put } u=x^2+1$
$\displaystyle \Rightarrow du=2x\,dx$
$\displaystyle \Rightarrow I_2=\frac{1}{2}\int\frac{1}{u}\,du$
$\displaystyle =\frac{1}{2}\log|u|+c_2$
$\displaystyle =\frac{1}{2}\log|x^2+1|+c_2\qquad (3)$
$\displaystyle \text{Now, } I_3=\int\frac{1}{x^2+1}\,dx$
$\displaystyle =\tan^{-1}x+c_3\qquad (4)$
$\displaystyle \text{From (1), (2), (3) and (4)}$
$\displaystyle I=\frac{1}{2}\log|x-1|-\frac{1}{2}\cdot\frac{1}{2}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$
$\displaystyle =\frac{1}{2}\log|x-1|-\frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$
$\displaystyle \therefore \int \frac{x}{(x^2+1)(x-1)}\,dx=\frac{1}{2}\log|x-1|-\frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 17: }~\int \frac{1}{(x-1)(x+1)(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{(x-1)(x+1)(x+2)}$
$\displaystyle \text{Let } \frac{1}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}$
$\displaystyle \Rightarrow \frac{1}{(x-1)(x+1)(x+2)}=\frac{A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)}{(x-1)(x+1)(x+2)}$
$\displaystyle \Rightarrow 1=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)$
$\displaystyle \text{Putting } x-1=0 \text{ or } x=1$
$\displaystyle 1=A(1+1)(1+2)+B(0)+C(0)$
$\displaystyle \Rightarrow 1=6A$
$\displaystyle \Rightarrow A=\frac{1}{6}$
$\displaystyle \text{Putting } x+1=0 \text{ or } x=-1$
$\displaystyle 1=A(0)+B(-2)(1)+C(0)$
$\displaystyle \Rightarrow B=-\frac{1}{2}$
$\displaystyle \text{Putting } x+2=0 \text{ or } x=-2$
$\displaystyle 1=A(0)+B(0)+C(-3)(-1)$
$\displaystyle \Rightarrow 1=3C$
$\displaystyle \Rightarrow C=\frac{1}{3}$
$\displaystyle \Rightarrow I=\frac{1}{6}\int\frac{dx}{x-1}-\frac{1}{2}\int\frac{dx}{x+1}+\frac{1}{3}\int\frac{dx}{x+2}$
$\displaystyle =\frac{1}{6}\log|x-1|-\frac{1}{2}\log|x+1|+\frac{1}{3}\log|x+2|+C$
$\displaystyle =\frac{1}{6}\log|x-1|-\frac{3}{6}\log|x+1|+\frac{2}{6}\log|x+2|+C$
$\displaystyle =\frac{1}{6}\log\left|\frac{(x-1)(x+2)^2}{(x+1)^3}\right|+C$

$\displaystyle \textbf{Question 18: }~\int \frac{x^2}{(x^2+4)(x^2+9)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x^2}{(x^2+4)(x^2+9)}\,dx$
$\displaystyle \text{We express } \frac{x^2}{(x^2+4)(x^2+9)}=\frac{A}{x^2+4}+\frac{B}{x^2+9}$
$\displaystyle \Rightarrow x^2=A(x^2+9)+B(x^2+4)$
$\displaystyle \Rightarrow x^2=(A+B)x^2+(9A+4B)$
$\displaystyle \text{Equating coefficients of } x^2 \text{ and constants}$
$\displaystyle A+B=1$
$\displaystyle 9A+4B=0$
$\displaystyle \Rightarrow A=-\frac{4}{5},\; B=\frac{9}{5}$
$\displaystyle \Rightarrow I=\int\left(-\frac{4}{5}\cdot\frac{1}{x^2+4}+\frac{9}{5}\cdot\frac{1}{x^2+9}\right)dx$
$\displaystyle =-\frac{4}{5}\int\frac{1}{x^2+4}\,dx+\frac{9}{5}\int\frac{1}{x^2+9}\,dx$
$\displaystyle =-\frac{4}{5}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{9}{5}\cdot\frac{1}{3}\tan^{-1}\frac{x}{3}+C$
$\displaystyle =-\frac{2}{5}\tan^{-1}\frac{x}{2}+\frac{3}{5}\tan^{-1}\frac{x}{3}+C$
$\displaystyle \therefore \int \frac{x^2}{(x^2+4)(x^2+9)}\,dx=-\frac{2}{5}\tan^{-1}\frac{x}{2}+\frac{3}{5}\tan^{-1}\frac{x}{3}+C$

$\displaystyle \textbf{Question 19: }~\int \frac{5x^2-1}{x(x-1)(x+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{5x^2-1}{x(x-1)(x+1)}\,dx$
$\displaystyle \text{Let } \frac{5x^2-1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$
$\displaystyle \Rightarrow 5x^2-1=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)$
$\displaystyle \Rightarrow 5x^2-1=A(x^2-1)+Bx(x+1)+Cx(x-1)$
$\displaystyle \text{Putting } x=1$
$\displaystyle 5(1)^2-1=A(0)+B(1)(2)+C(0)$
$\displaystyle \Rightarrow 4=2B$
$\displaystyle \Rightarrow B=2$
$\displaystyle \text{Putting } x=0$
$\displaystyle 5(0)^2-1=A(-1)+B(0)+C(0)$
$\displaystyle \Rightarrow -1=-A$
$\displaystyle \Rightarrow A=1$
$\displaystyle \text{Putting } x=-1$
$\displaystyle 5(1)-1=A(0)+B(0)+C(-1)(-2)$
$\displaystyle \Rightarrow 4=2C$
$\displaystyle \Rightarrow C=2$
$\displaystyle \Rightarrow \frac{5x^2-1}{x(x-1)(x+1)}=\frac{1}{x}+\frac{2}{x-1}+\frac{2}{x+1}$
$\displaystyle \Rightarrow I=\int\frac{dx}{x}+2\int\frac{dx}{x-1}+2\int\frac{dx}{x+1}$
$\displaystyle =\log|x|+2\log|x-1|+2\log|x+1|+C$
$\displaystyle =\log|x|+2\log|x^2-1|+C$
$\displaystyle =\log\left|x(x^2-1)^2\right|+C$

$\displaystyle \textbf{Question 20: }~\int \frac{x^2+6x-8}{x^3-4x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2+6x-8}{x^3-4x}\,dx$
$\displaystyle =\int \frac{x^2+6x-8}{x(x^2-4)}\,dx$
$\displaystyle =\int \frac{x^2+6x-8}{x(x-2)(x+2)}\,dx$
$\displaystyle \text{Let } \frac{x^2+6x-8}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2}$
$\displaystyle \Rightarrow x^2+6x-8=A(x-2)(x+2)+Bx(x+2)+Cx(x-2)$
$\displaystyle \Rightarrow x^2+6x-8=A(x^2-4)+B(x^2+2x)+C(x^2-2x)$
$\displaystyle \text{Putting } x=2$
$\displaystyle 4+12-8=A(0)+B(2)(4)+C(0)$
$\displaystyle \Rightarrow 8=8B$
$\displaystyle \Rightarrow B=1$
$\displaystyle \text{Putting } x=-2$
$\displaystyle 4-12-8=A(0)+B(0)+C(-2)(-4)$
$\displaystyle \Rightarrow -16=8C$
$\displaystyle \Rightarrow C=-2$
$\displaystyle \text{Putting } x=0$
$\displaystyle -8=A(-4)+B(0)+C(0)$
$\displaystyle \Rightarrow A=2$
$\displaystyle \Rightarrow \frac{x^2+6x-8}{x(x-2)(x+2)}=\frac{2}{x}+\frac{1}{x-2}-\frac{2}{x+2}$
$\displaystyle \Rightarrow I=\int\frac{2}{x}\,dx+\int\frac{1}{x-2}\,dx-2\int\frac{1}{x+2}\,dx$
$\displaystyle =2\log|x|+\log|x-2|-2\log|x+2|+C$
$\displaystyle =\log x^2+\log|x-2|-\log|x+2|^2+C$
$\displaystyle =\log\left|\frac{x^2(x-2)}{(x+2)^2}\right|+C$

$\displaystyle \textbf{Question 21: }~\int \frac{x^2+1}{(2x+1)(x^2-1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2+1}{(2x+1)(x^2-1)}\,dx$
$\displaystyle =\int \frac{x^2+1}{(2x+1)(x-1)(x+1)}\,dx$
$\displaystyle \text{Let } \frac{x^2+1}{(2x+1)(x-1)(x+1)}=\frac{A}{2x+1}+\frac{B}{x-1}+\frac{C}{x+1}$
$\displaystyle \Rightarrow x^2+1=A(x^2-1)+B(2x+1)(x+1)+C(2x+1)(x-1)$
$\displaystyle \text{Putting } x=1$
$\displaystyle 1^2+1=A(0)+B(3)(2)+C(0)$
$\displaystyle \Rightarrow 2=6B$
$\displaystyle \Rightarrow B=\frac{1}{3}$
$\displaystyle \text{Putting } x=-1$
$\displaystyle 1^2+1=A(0)+B(0)+C(-1)(-2)$
$\displaystyle \Rightarrow 2=2C$
$\displaystyle \Rightarrow C=1$
$\displaystyle \text{Putting } 2x+1=0 \text{ or } x=-\frac{1}{2}$
$\displaystyle \left(-\frac{1}{2}\right)^2+1=A\left(\frac{1}{4}-1\right)$
$\displaystyle \Rightarrow \frac{5}{4}=A\left(-\frac{3}{4}\right)$
$\displaystyle \Rightarrow A=-\frac{5}{3}$
$\displaystyle \Rightarrow \frac{x^2+1}{(2x+1)(x-1)(x+1)}=-\frac{5}{3(2x+1)}+\frac{1}{3(x-1)}+\frac{1}{x+1}$
$\displaystyle \Rightarrow I=-\frac{5}{3}\int\frac{dx}{2x+1}+\frac{1}{3}\int\frac{dx}{x-1}+\int\frac{dx}{x+1}$
$\displaystyle =-\frac{5}{3}\cdot\frac{1}{2}\log|2x+1|+\frac{1}{3}\log|x-1|+\log|x+1|+C$
$\displaystyle =-\frac{5}{6}\log|2x+1|+\frac{1}{3}\log|x-1|+\log|x+1|+C$

$\displaystyle \textbf{Question 22: }~\int \frac{1}{x\{6(\log x)^2+7\log x+2\}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{x\{6(\log x)^2+7\log x+2\}}$
$\displaystyle \text{Put } t=\log x$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt$
$\displaystyle \Rightarrow I=\int \frac{dt}{6t^2+7t+2}$
$\displaystyle =\int \frac{dt}{(3t+2)(2t+1)}$
$\displaystyle \text{Let } \frac{1}{(3t+2)(2t+1)}=\frac{A}{3t+2}+\frac{B}{2t+1}$
$\displaystyle \Rightarrow \frac{1}{(3t+2)(2t+1)}=\frac{A(2t+1)+B(3t+2)}{(3t+2)(2t+1)}$
$\displaystyle \Rightarrow 1=A(2t+1)+B(3t+2)$
$\displaystyle \text{Putting } 2t+1=0$
$\displaystyle \Rightarrow t=-\frac{1}{2}$
$\displaystyle 1=A(0)+B\left(3\left(-\frac{1}{2}\right)+2\right)$
$\displaystyle \Rightarrow 1=B\left(\frac{1}{2}\right)$
$\displaystyle \Rightarrow B=2$
$\displaystyle \text{Putting } 3t+2=0$
$\displaystyle \Rightarrow t=-\frac{2}{3}$
$\displaystyle 1=A\left(2\left(-\frac{2}{3}\right)+1\right)+0$
$\displaystyle \Rightarrow 1=A\left(-\frac{1}{3}\right)$
$\displaystyle \Rightarrow A=-3$
$\displaystyle \Rightarrow I=\int\left(-\frac{3}{3t+2}+\frac{2}{2t+1}\right)dt$
$\displaystyle =-\log|3t+2|+\log|2t+1|+C$
$\displaystyle =\log\left|\frac{2t+1}{3t+2}\right|+C$
$\displaystyle =\log\left|\frac{2\log x+1}{3\log x+2}\right|+C$

$\displaystyle \textbf{Question 23: }~\int \frac{1}{x(x^n+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{x(x^n+1)}$
$\displaystyle =\int \frac{x^{n-1}\,dx}{x^n(x^n+1)}$
$\displaystyle =\int \frac{x^{n-1}\,dx}{x^n(x^n+1)}$
$\displaystyle \text{Put } x^n=t$
$\displaystyle \Rightarrow nx^{n-1}\,dx=dt$
$\displaystyle \Rightarrow x^{n-1}\,dx=\frac{dt}{n}$
$\displaystyle \Rightarrow I=\frac{1}{n}\int \frac{dt}{t(t+1)}$
$\displaystyle \text{Let } \frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}$
$\displaystyle \Rightarrow \frac{1}{t(t+1)}=\frac{A(t+1)+Bt}{t(t+1)}$
$\displaystyle \Rightarrow 1=A(t+1)+Bt$
$\displaystyle \text{Putting } t+1=0$
$\displaystyle \Rightarrow t=-1$
$\displaystyle 1=A(0)+B(-1)$
$\displaystyle \Rightarrow B=-1$
$\displaystyle \text{Putting } t=0$
$\displaystyle 1=A(1)+B(0)$
$\displaystyle \Rightarrow A=1$
$\displaystyle \Rightarrow I=\frac{1}{n}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt$
$\displaystyle =\frac{1}{n}\log|t|-\frac{1}{n}\log|t+1|+C$
$\displaystyle =\frac{1}{n}\log\left|\frac{t}{t+1}\right|+C$
$\displaystyle =\frac{1}{n}\log\left|\frac{x^n}{x^n+1}\right|+C$

$\displaystyle \textbf{Question 24: }~\int \frac{x}{(x^2-a^2)(x^2-b^2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x\,dx}{(x^2-a^2)(x^2-b^2)}$
$\displaystyle \text{Put } t=x^2$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int \frac{dt}{(t-a^2)(t-b^2)}$
$\displaystyle \text{Let } \frac{1}{(t-a^2)(t-b^2)}=\frac{A}{t-a^2}+\frac{B}{t-b^2}$
$\displaystyle \Rightarrow \frac{1}{(t-a^2)(t-b^2)}=\frac{A(t-b^2)+B(t-a^2)}{(t-a^2)(t-b^2)}$
$\displaystyle \Rightarrow 1=A(t-b^2)+B(t-a^2)$
$\displaystyle \text{Putting } t=b^2$
$\displaystyle 1=A(0)+B(b^2-a^2)$
$\displaystyle \Rightarrow B=\frac{1}{b^2-a^2}$
$\displaystyle \text{Putting } t=a^2$
$\displaystyle 1=A(a^2-b^2)+B(0)$
$\displaystyle \Rightarrow A=\frac{1}{a^2-b^2}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\left(\frac{1}{a^2-b^2}\cdot\frac{1}{t-a^2}+\frac{1}{b^2-a^2}\cdot\frac{1}{t-b^2}\right)dt$
$\displaystyle =\frac{1}{2(a^2-b^2)}\int\frac{dt}{t-a^2}-\frac{1}{2(a^2-b^2)}\int\frac{dt}{t-b^2}$
$\displaystyle =\frac{1}{2(a^2-b^2)}\log|t-a^2|-\frac{1}{2(a^2-b^2)}\log|t-b^2|+C$
$\displaystyle =\frac{1}{2(a^2-b^2)}\log\left|\frac{t-a^2}{t-b^2}\right|+C$
$\displaystyle =\frac{1}{2(a^2-b^2)}\log\left|\frac{x^2-a^2}{x^2-b^2}\right|+C$

$\displaystyle \textbf{Question 25: }~\int \frac{x^2+1}{(x^2+4)(x^2+25)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x^2+1}{(x^2+4)(x^2+25)}\,dx$
$\displaystyle \text{We express } \frac{x^2+1}{(x^2+4)(x^2+25)}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+25}$
$\displaystyle \Rightarrow x^2+1=(Ax+B)(x^2+25)+(Cx+D)(x^2+4)$
$\displaystyle \Rightarrow x^2+1=(A+C)x^3+(B+D)x^2+(25A+4C)x+(25B+4D)$
$\displaystyle \text{Equating coefficients of } x^3,x^2,x \text{ and constants}$
$\displaystyle A+C=0$
$\displaystyle B+D=1$
$\displaystyle 25A+4C=0$
$\displaystyle 25B+4D=1$
$\displaystyle \Rightarrow A=0,\; C=0,\; B=-\frac{1}{7},\; D=\frac{8}{7}$
$\displaystyle \Rightarrow I=\int\left(-\frac{1}{7}\cdot\frac{1}{x^2+4}+\frac{8}{7}\cdot\frac{1}{x^2+25}\right)dx$
$\displaystyle =-\frac{1}{7}\int\frac{1}{x^2+4}\,dx+\frac{8}{7}\int\frac{1}{x^2+25}\,dx$
$\displaystyle =-\frac{1}{7}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{8}{7}\cdot\frac{1}{5}\tan^{-1}\frac{x}{5}+C$
$\displaystyle =-\frac{1}{14}\tan^{-1}\frac{x}{2}+\frac{8}{35}\tan^{-1}\frac{x}{5}+C$
$\displaystyle \therefore \int \frac{x^2+1}{(x^2+4)(x^2+25)}\,dx=-\frac{1}{14}\tan^{-1}\frac{x}{2}+\frac{8}{35}\tan^{-1}\frac{x}{5}+C$

$\displaystyle \textbf{Question 26: }~\int \frac{x^3+x+1}{x^2-1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First do division: }\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2x+1}{x^{2}-1}.$
$\displaystyle \int \frac{x^{3}+x+1}{x^{2}-1}\,dx=\int x\,dx+\int \frac{2x+1}{x^{2}-1}\,dx.$
$\displaystyle x^{2}-1=(x-1)(x+1).$
$\displaystyle \frac{2x+1}{x^{2}-1}=\frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.$
$\displaystyle 2x+1=A(x+1)+B(x-1)=(A+B)x+(A-B).$
$\displaystyle \text{Comparing coefficients: }A+B=2,\;A-B=1.$
$\displaystyle \Rightarrow 2A=3\;\Rightarrow\;A=\frac{3}{2},\quad B=\frac{1}{2}.$
$\displaystyle \therefore \frac{2x+1}{x^{2}-1}=\frac{3}{2}\cdot\frac{1}{x-1}+\frac{1}{2}\cdot\frac{1}{x+1}.$
$\displaystyle \int \frac{x^{3}+x+1}{x^{2}-1}\,dx=\int x\,dx+\frac{3}{2}\int \frac{1}{x-1}\,dx+\frac{1}{2}\int \frac{1}{x+1}\,dx.$
$\displaystyle =\frac{x^{2}}{2}+\frac{3}{2}\log|x-1|+\frac{1}{2}\log|x+1|+C.$

$\displaystyle \textbf{Question 27: }~\int \frac{3x-2}{(x+1)^2(x+3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{3x-2}{(x+1)^2(x+3)}\,dx$
$\displaystyle \text{We express } \frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$
$\displaystyle \Rightarrow 3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2$
$\displaystyle \Rightarrow 3x-2=A(x^2+4x+3)+B(x+3)+C(x^2+2x+1)$
$\displaystyle \Rightarrow 3x-2=(A+C)x^2+(4A+B+2C)x+(3A+3B+C)$
$\displaystyle \text{Equating coefficients of } x^2,x \text{ and constants}$
$\displaystyle A+C=0$
$\displaystyle 4A+B+2C=3$
$\displaystyle 3A+3B+C=-2$
$\displaystyle \Rightarrow A=\frac{11}{4},\; B=-\frac{5}{2},\; C=-\frac{11}{4}$
$\displaystyle \Rightarrow I=\int\left(\frac{11}{4}\cdot\frac{1}{x+1}-\frac{5}{2}\cdot\frac{1}{(x+1)^2}-\frac{11}{4}\cdot\frac{1}{x+3}\right)dx$
$\displaystyle =\frac{11}{4}\int\frac{1}{x+1}\,dx-\frac{5}{2}\int\frac{1}{(x+1)^2}\,dx-\frac{11}{4}\int\frac{1}{x+3}\,dx$
$\displaystyle =\frac{11}{4}\log|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\log|x+3|+C$
$\displaystyle \therefore \int \frac{3x-2}{(x+1)^2(x+3)}\,dx=\frac{11}{4}\log|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\log|x+3|+C$

$\displaystyle \textbf{Question 28: }~\int \frac{2x+1}{(x+2)(x-3)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{2x+1}{(x+2)(x-3)^2}\,dx$
$\displaystyle \text{Let } \frac{2x+1}{(x+2)(x-3)^2}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{(x-3)^2}$
$\displaystyle \Rightarrow \frac{2x+1}{(x+2)(x-3)^2}=\frac{A(x-3)^2+B(x+2)(x-3)+C(x+2)}{(x+2)(x-3)^2}$
$\displaystyle \Rightarrow 2x+1=A(x^2-6x+9)+B(x^2-x-6)+C(x+2)$
$\displaystyle \Rightarrow 2x+1=(A+B)x^2+(-6A-B+C)x+(9A-6B+2C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle -6A-B+C=2\qquad (2)$
$\displaystyle 9A-6B+2C=1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=-\frac{3}{25},\; B=\frac{3}{25},\; C=\frac{7}{5}$
$\displaystyle \Rightarrow \frac{2x+1}{(x+2)(x-3)^2}=-\frac{3}{25}\cdot\frac{1}{x+2}+\frac{3}{25}\cdot\frac{1}{x-3}+\frac{7}{5}\cdot\frac{1}{(x-3)^2}$
$\displaystyle \Rightarrow I=-\frac{3}{25}\int\frac{dx}{x+2}+\frac{3}{25}\int\frac{dx}{x-3}+\frac{7}{5}\int\frac{dx}{(x-3)^2}$
$\displaystyle =-\frac{3}{25}\log|x+2|+\frac{3}{25}\log|x-3|+\frac{7}{5}\int (x-3)^{-2}\,dx$
$\displaystyle =-\frac{3}{25}\log|x+2|+\frac{3}{25}\log|x-3|-\frac{7}{5(x-3)}+C$

$\displaystyle \textbf{Question 29: }~\int \frac{x^2+1}{(x-2)^2(x+3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2+1}{(x-2)^2(x+3)}\,dx$
$\displaystyle \text{Let } \frac{x^2+1}{(x-2)^2(x+3)}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+3}$
$\displaystyle \Rightarrow \frac{x^2+1}{(x-2)^2(x+3)}=\frac{A(x-2)(x+3)+B(x+3)+C(x-2)^2}{(x-2)^2(x+3)}$
$\displaystyle \Rightarrow x^2+1=A(x^2+x-6)+B(x+3)+C(x^2-4x+4)$
$\displaystyle \Rightarrow x^2+1=(A+C)x^2+(A+B-4C)x+(-6A+3B+4C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+C=1\qquad (1)$
$\displaystyle A+B-4C=0\qquad (2)$
$\displaystyle -6A+3B+4C=1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{3}{5},\; B=1,\; C=\frac{2}{5}$
$\displaystyle \Rightarrow I=\frac{3}{5}\int\frac{dx}{x-2}+\int\frac{dx}{(x-2)^2}+\frac{2}{5}\int\frac{dx}{x+3}$
$\displaystyle =\frac{3}{5}\log|x-2|+\int (x-2)^{-2}\,dx+\frac{2}{5}\log|x+3|+C$
$\displaystyle =\frac{3}{5}\log|x-2|-\frac{1}{x-2}+\frac{2}{5}\log|x+3|+C$

$\displaystyle \textbf{Question 30: }~\int \frac{x}{(x-1)^2(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x}{(x-1)^2(x+2)}\,dx$
$\displaystyle \text{Let } \frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}$
$\displaystyle \Rightarrow \frac{x}{(x-1)^2(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}$
$\displaystyle \Rightarrow x=A(x^2+x-2)+B(x+2)+C(x^2-2x+1)$
$\displaystyle \Rightarrow x=(A+C)x^2+(A+B-2C)x+(-2A+2B+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+C=0\qquad (1)$
$\displaystyle A+B-2C=1\qquad (2)$
$\displaystyle -2A+2B+C=0\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{2}{9},\; B=\frac{1}{3},\; C=-\frac{2}{9}$
$\displaystyle \Rightarrow \frac{x}{(x-1)^2(x+2)}=\frac{2}{9}\cdot\frac{1}{x-1}+\frac{1}{3}\cdot\frac{1}{(x-1)^2}-\frac{2}{9}\cdot\frac{1}{x+2}$
$\displaystyle \Rightarrow I=\frac{2}{9}\int\frac{dx}{x-1}+\frac{1}{3}\int\frac{dx}{(x-1)^2}-\frac{2}{9}\int\frac{dx}{x+2}$
$\displaystyle =\frac{2}{9}\log|x-1|+\frac{1}{3}\int (x-1)^{-2}\,dx-\frac{2}{9}\log|x+2|+C$
$\displaystyle =\frac{2}{9}\log|x-1|-\frac{1}{3(x-1)}-\frac{2}{9}\log|x+2|+C$
$\displaystyle =\frac{2}{9}\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C$

$\displaystyle \textbf{Question 31: }~\int \frac{x^2}{(x-1)(x+1)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2}{(x-1)(x+1)^2}\,dx$
$\displaystyle \text{Let } \frac{x^2}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
$\displaystyle \Rightarrow \frac{x^2}{(x-1)(x+1)^2}=\frac{A(x+1)^2+B(x+1)(x-1)+C(x-1)}{(x-1)(x+1)^2}$
$\displaystyle \Rightarrow x^2=A(x^2+2x+1)+B(x^2-1)+C(x-1)$
$\displaystyle \Rightarrow x^2=(A+B)x^2+(2A+C)x+(A-B-C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=1\qquad (1)$
$\displaystyle 2A+C=0\qquad (2)$
$\displaystyle A-B-C=0\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{1}{4},\; B=\frac{3}{4},\; C=-\frac{1}{2}$
$\displaystyle \Rightarrow \frac{x^2}{(x-1)(x+1)^2}=\frac{1}{4}\cdot\frac{1}{x-1}+\frac{3}{4}\cdot\frac{1}{x+1}-\frac{1}{2}\cdot\frac{1}{(x+1)^2}$
$\displaystyle \Rightarrow I=\frac{1}{4}\int\frac{dx}{x-1}+\frac{3}{4}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{(x+1)^2}$
$\displaystyle =\frac{1}{4}\log|x-1|+\frac{3}{4}\log|x+1|-\frac{1}{2}\int (x+1)^{-2}\,dx$
$\displaystyle =\frac{1}{4}\log|x-1|+\frac{3}{4}\log|x+1|+\frac{1}{2(x+1)}+C$

$\displaystyle \textbf{Question 32: }~\int \frac{x^2+x-1}{(x+1)^2(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2+x-1}{(x+1)^2(x+2)}\,dx$
$\displaystyle \text{Let } \frac{x^2+x-1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$
$\displaystyle \Rightarrow \frac{x^2+x-1}{(x+1)^2(x+2)}=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^2}{(x+1)^2(x+2)}$
$\displaystyle \Rightarrow x^2+x-1=A(x^2+3x+2)+B(x+2)+C(x^2+2x+1)$
$\displaystyle \Rightarrow x^2+x-1=(A+C)x^2+(3A+B+2C)x+(2A+2B+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+C=1\qquad (1)$
$\displaystyle 3A+B+2C=1\qquad (2)$
$\displaystyle 2A+2B+C=-1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=0,\; B=-1,\; C=1$
$\displaystyle \Rightarrow \frac{x^2+x-1}{(x+1)^2(x+2)}=-\frac{1}{(x+1)^2}+\frac{1}{x+2}$
$\displaystyle \Rightarrow I=-\int\frac{dx}{(x+1)^2}+\int\frac{dx}{x+2}$
$\displaystyle =-\int (x+1)^{-2}\,dx+\log|x+2|+C$
$\displaystyle =\frac{1}{x+1}+\log|x+2|+C$

$\displaystyle \textbf{Question 33: }~\int \frac{2x^2+7x-3}{x^2(2x+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{2x^2+7x-3}{x^2(2x+1)}\,dx$
$\displaystyle \text{Let } \frac{2x^2+7x-3}{x^2(2x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{2x+1}$
$\displaystyle \Rightarrow \frac{2x^2+7x-3}{x^2(2x+1)}=\frac{A x(2x+1)+B(2x+1)+C x^2}{x^2(2x+1)}$
$\displaystyle \Rightarrow 2x^2+7x-3=A(2x^2+x)+B(2x+1)+C x^2$
$\displaystyle \Rightarrow 2x^2+7x-3=(2A+C)x^2+(A+2B)x+B$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle 2A+C=2\qquad (1)$
$\displaystyle A+2B=7\qquad (2)$
$\displaystyle B=-3\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=13,\; B=-3,\; C=-24$
$\displaystyle \Rightarrow \frac{2x^2+7x-3}{x^2(2x+1)}=\frac{13}{x}-\frac{3}{x^2}-\frac{24}{2x+1}$
$\displaystyle \Rightarrow I=13\int\frac{dx}{x}-3\int x^{-2}\,dx-24\int\frac{dx}{2x+1}$
$\displaystyle =13\log|x|+\frac{3}{x}-24\cdot\frac{1}{2}\log|2x+1|+C$
$\displaystyle =13\log|x|+\frac{3}{x}-12\log|2x+1|+C$

$\displaystyle \textbf{Question 34: }~\int \frac{5x^2+20x+6}{x^3+2x^2+x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{5x^2+20x+6}{x^3+2x^2+x}\,dx$
$\displaystyle =\int \frac{5x^2+20x+6}{x(x^2+2x+1)}\,dx$
$\displaystyle =\int \frac{5x^2+20x+6}{x(x+1)^2}\,dx$
$\displaystyle \text{Let } \frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
$\displaystyle \Rightarrow \frac{5x^2+20x+6}{x(x+1)^2}=\frac{A(x+1)^2+B x(x+1)+C x}{x(x+1)^2}$
$\displaystyle \Rightarrow 5x^2+20x+6=A(x^2+2x+1)+B(x^2+x)+Cx$
$\displaystyle \Rightarrow 5x^2+20x+6=(A+B)x^2+(2A+B+C)x+A$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=5\qquad (1)$
$\displaystyle 2A+B+C=20\qquad (2)$
$\displaystyle A=6\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=6,\; B=-1,\; C=9$
$\displaystyle \Rightarrow \frac{5x^2+20x+6}{x(x+1)^2}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2}$
$\displaystyle \Rightarrow I=6\int\frac{dx}{x}-\int\frac{dx}{x+1}+9\int\frac{dx}{(x+1)^2}$
$\displaystyle =6\log|x|-\log|x+1|+9\int (x+1)^{-2}\,dx$
$\displaystyle =6\log|x|-\log|x+1|-\frac{9}{x+1}+C$

$\displaystyle \textbf{Question 35: }~\int \frac{18}{(x+2)(x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{18}{(x+2)(x^2+4)}\,dx$
$\displaystyle \text{Let } \frac{18}{(x+2)(x^2+4)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+4}$
$\displaystyle \Rightarrow \frac{18}{(x+2)(x^2+4)}=\frac{A(x^2+4)+(Bx+C)(x+2)}{(x+2)(x^2+4)}$
$\displaystyle \Rightarrow 18=A(x^2+4)+Bx^2+2Bx+Cx+2C$
$\displaystyle \Rightarrow 18=(A+B)x^2+(2B+C)x+(4A+2C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle 2B+C=0\qquad (2)$
$\displaystyle 4A+2C=18\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{9}{4},\; B=-\frac{9}{4},\; C=\frac{9}{2}$
$\displaystyle \Rightarrow \frac{18}{(x+2)(x^2+4)}=\frac{9}{4}\cdot\frac{1}{x+2}-\frac{9}{4}\cdot\frac{x}{x^2+4}+\frac{9}{2}\cdot\frac{1}{x^2+4}$
$\displaystyle \Rightarrow I=\frac{9}{4}\int\frac{dx}{x+2}-\frac{9}{4}\int\frac{x\,dx}{x^2+4}+\frac{9}{2}\int\frac{dx}{x^2+4}$
$\displaystyle \text{Put } t=x^2+4$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+4}=\frac{1}{2}\log|x^2+4|$
$\displaystyle \Rightarrow I=\frac{9}{4}\log|x+2|-\frac{9}{8}\log|x^2+4|+\frac{9}{2}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+C$
$\displaystyle =\frac{9}{4}\log|x+2|-\frac{9}{8}\log|x^2+4|+\frac{9}{4}\tan^{-1}\frac{x}{2}+C$

$\displaystyle \textbf{Question 36: }~\int \frac{5}{(x^2+1)(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{5\,dx}{(x^2+1)(x+2)}$
$\displaystyle \text{Let } \frac{5}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow \frac{5}{(x+2)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x+2)}{(x+2)(x^2+1)}$
$\displaystyle \Rightarrow 5=A(x^2+1)+Bx^2+2Bx+Cx+2C$
$\displaystyle \Rightarrow 5=(A+B)x^2+(2B+C)x+(A+2C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle 2B+C=0\qquad (2)$
$\displaystyle A+2C=5\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=1,\; B=-1,\; C=2$
$\displaystyle \Rightarrow \frac{5}{(x+2)(x^2+1)}=\frac{1}{x+2}-\frac{x}{x^2+1}+\frac{2}{x^2+1}$
$\displaystyle \Rightarrow I=\int\frac{dx}{x+2}-\int\frac{x\,dx}{x^2+1}+2\int\frac{dx}{x^2+1}$
$\displaystyle \text{Put } t=x^2+1$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac{1}{2}\log|x^2+1|$
$\displaystyle \Rightarrow I=\log|x+2|-\frac{1}{2}\log|x^2+1|+2\tan^{-1}x+C$

$\displaystyle \textbf{Question 37: }~\int \frac{x}{(x+1)(x^2+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x\,dx}{(x+1)(x^2+1)}$
$\displaystyle \text{Let } \frac{x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow \frac{x}{(x+1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}$
$\displaystyle \Rightarrow x=A(x^2+1)+Bx^2+Bx+Cx+C$
$\displaystyle \Rightarrow x=(A+B)x^2+(B+C)x+(A+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle B+C=1\qquad (2)$
$\displaystyle A+C=0\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=-\frac{1}{2},\; B=\frac{1}{2},\; C=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{x}{(x+1)(x^2+1)}=-\frac{1}{2}\cdot\frac{1}{x+1}+\frac{1}{2}\cdot\frac{x}{x^2+1}+\frac{1}{2}\cdot\frac{1}{x^2+1}$
$\displaystyle \Rightarrow I=-\frac{1}{2}\int\frac{dx}{x+1}+\frac{1}{2}\int\frac{x\,dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x^2+1}$
$\displaystyle \text{Put } t=x^2+1$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac{1}{2}\log|t|=\frac{1}{2}\log|x^2+1|$
$\displaystyle \Rightarrow I=-\frac{1}{2}\log|x+1|+\frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 38: }~\int \frac{1}{1+x+x^2+x^3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{1+x+x^2+x^3}$
$\displaystyle =\int \frac{dx}{(1+x)+x^2(1+x)}$
$\displaystyle =\int \frac{dx}{(x+1)(x^2+1)}$
$\displaystyle \text{Let } \frac{1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow \frac{1}{(x+1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}$
$\displaystyle \Rightarrow 1=A(x^2+1)+Bx^2+Bx+Cx+C$
$\displaystyle \Rightarrow 1=(A+B)x^2+(B+C)x+(A+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle B+C=0\qquad (2)$
$\displaystyle A+C=1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{1}{2},\; B=-\frac{1}{2},\; C=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1}{(x+1)(x^2+1)}=\frac{1}{2}\cdot\frac{1}{x+1}-\frac{1}{2}\cdot\frac{x}{x^2+1}+\frac{1}{2}\cdot\frac{1}{x^2+1}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{x\,dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x^2+1}$
$\displaystyle \text{Put } t=x^2+1$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac{1}{2}\log|x^2+1|$
$\displaystyle \Rightarrow I=\frac{1}{2}\log|x+1|-\frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 39: }~\int \frac{1}{(x+1)^2(x^2+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{(x+1)^2(x^2+1)}$
$\displaystyle \text{Let } \frac{1}{(x+1)^2(x^2+1)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+1}$
$\displaystyle \Rightarrow \frac{1}{(x+1)^2(x^2+1)}=\frac{A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2}{(x+1)^2(x^2+1)}$
$\displaystyle \Rightarrow 1=A(x^3+x^2+x+1)+B(x^2+1)+(Cx+D)(x^2+2x+1)$
$\displaystyle \Rightarrow 1=A(x^3+x^2+x+1)+B(x^2+1)+Cx^3+2Cx^2+Cx+Dx^2+2Dx+D$
$\displaystyle \Rightarrow 1=(A+C)x^3+(A+B+2C+D)x^2+(A+C+2D)x+(A+B+D)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+C=0\qquad (1)$
$\displaystyle A+B+2C+D=0\qquad (2)$
$\displaystyle A+C+2D=0\qquad (3)$
$\displaystyle A+B+D=1\qquad (4)$
$\displaystyle \text{Solving (1), (2), (3) and (4), we get}$
$\displaystyle A=\frac{1}{2},\; B=\frac{1}{2},\; C=-\frac{1}{2},\; D=0$
$\displaystyle \Rightarrow \frac{1}{(x+1)^2(x^2+1)}=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^2}-\frac{1}{2}\cdot\frac{x}{x^2+1}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\frac{dx}{x+1}+\frac{1}{2}\int\frac{dx}{(x+1)^2}-\frac{1}{2}\int\frac{x\,dx}{x^2+1}$
$\displaystyle \text{Put } t=x^2+1$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac{1}{2}\log|t|=\frac{1}{2}\log|x^2+1|$
$\displaystyle \Rightarrow I=\frac{1}{2}\log|x+1|-\frac{1}{2(x+1)}-\frac{1}{4}\log|x^2+1|+C$

$\displaystyle \textbf{Question 40: }~\int \frac{2x}{x^3-1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{2x\,dx}{x^3-1}$
$\displaystyle =\int \frac{2x\,dx}{(x-1)(x^2+x+1)}$
$\displaystyle \text{Let } \frac{2x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$\displaystyle \Rightarrow \frac{2x}{(x-1)(x^2+x+1)}=\frac{A(x^2+x+1)+(Bx+C)(x-1)}{(x-1)(x^2+x+1)}$
$\displaystyle \Rightarrow 2x=A(x^2+x+1)+Bx^2-Bx+Cx-C$
$\displaystyle \Rightarrow 2x=(A+B)x^2+(A-B+C)x+(A-C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle A-B+C=2\qquad (2)$
$\displaystyle A-C=0\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{2}{3},\; B=-\frac{2}{3},\; C=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{2x}{(x-1)(x^2+x+1)}=\frac{2}{3}\cdot\frac{1}{x-1}+\frac{2}{3}\cdot\frac{-x+1}{x^2+x+1}$
$\displaystyle \Rightarrow I=\frac{2}{3}\int\frac{dx}{x-1}+\frac{2}{3}\int\frac{-x+1}{x^2+x+1}\,dx$
$\displaystyle \text{Let } -x+1=a\frac{d}{dx}(x^2+x+1)+b$
$\displaystyle \Rightarrow -x+1=a(2x+1)+b$
$\displaystyle \text{Equating coefficients}$
$\displaystyle 2a=-1$
$\displaystyle \Rightarrow a=-\frac{1}{2}$
$\displaystyle a+b=1$
$\displaystyle \Rightarrow b=\frac{3}{2}$
$\displaystyle \Rightarrow I=\frac{2}{3}\int\frac{dx}{x-1}-\frac{1}{3}\int\frac{(2x+1)\,dx}{x^2+x+1}+\int\frac{dx}{x^2+x+1}$
$\displaystyle \text{Let } t=x^2+x+1$
$\displaystyle \Rightarrow dt=(2x+1)\,dx$
$\displaystyle \Rightarrow I=\frac{2}{3}\log|x-1|-\frac{1}{3}\log|t|+\int\frac{dx}{(x+\tfrac12)^2+\left(\tfrac{\sqrt3}{2}\right)^2}$
$\displaystyle =\frac{2}{3}\log|x-1|-\frac{1}{3}\log|x^2+x+1|+\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)+C$

$\displaystyle \textbf{Question 41: }~\int \frac{1}{(x^2+1)(x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{(x^2+1)(x^2+4)}$
$\displaystyle \text{Put } x^2=t$
$\displaystyle \Rightarrow \frac{1}{(x^2+1)(x^2+4)}=\frac{1}{(t+1)(t+4)}$
$\displaystyle \text{Let } \frac{1}{(t+1)(t+4)}=\frac{A}{t+1}+\frac{B}{t+4}$
$\displaystyle \Rightarrow \frac{1}{(t+1)(t+4)}=\frac{A(t+4)+B(t+1)}{(t+1)(t+4)}$
$\displaystyle \Rightarrow 1=A(t+4)+B(t+1)$
$\displaystyle \text{Putting } t+4=0$
$\displaystyle \Rightarrow t=-4$
$\displaystyle \Rightarrow 1=A\cdot 0+B(-3)$
$\displaystyle \Rightarrow B=-\frac{1}{3}$
$\displaystyle \text{Putting } t+1=0$
$\displaystyle \Rightarrow t=-1$
$\displaystyle \Rightarrow 1=A(3)+B\cdot 0$
$\displaystyle \Rightarrow A=\frac{1}{3}$
$\displaystyle \Rightarrow \frac{1}{(t+1)(t+4)}=\frac{1}{3(t+1)}-\frac{1}{3(t+4)}$
$\displaystyle \Rightarrow \frac{1}{(x^2+1)(x^2+4)}=\frac{1}{3(x^2+1)}-\frac{1}{3(x^2+4)}$
$\displaystyle \Rightarrow I=\frac{1}{3}\int\frac{dx}{x^2+1}-\frac{1}{3}\int\frac{dx}{x^2+4}$
$\displaystyle =\frac{1}{3}\tan^{-1}x-\frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C$
$\displaystyle =\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}\left(\frac{x}{2}\right)+C$

$\displaystyle \textbf{Question 42: }~\int \frac{x^2}{(x^2+1)(3x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2\,dx}{(x^2+1)(3x^2+4)}$
$\displaystyle \text{Put } x^2=t$
$\displaystyle \Rightarrow \frac{x^2}{(x^2+1)(3x^2+4)}=\frac{t}{(t+1)(3t+4)}$
$\displaystyle \text{Let } \frac{t}{(t+1)(3t+4)}=\frac{A}{t+1}+\frac{B}{3t+4}$
$\displaystyle \Rightarrow \frac{t}{(t+1)(3t+4)}=\frac{A(3t+4)+B(t+1)}{(t+1)(3t+4)}$
$\displaystyle \Rightarrow t=A(3t+4)+B(t+1)$
$\displaystyle \text{Putting } t+1=0$
$\displaystyle \Rightarrow t=-1$
$\displaystyle \Rightarrow -1=A(-3+4)+0$
$\displaystyle \Rightarrow A=-1$
$\displaystyle \text{Putting } 3t+4=0$
$\displaystyle \Rightarrow t=-\frac{4}{3}$
$\displaystyle \Rightarrow -\frac{4}{3}=0+B\left(-\frac{4}{3}+1\right)$
$\displaystyle \Rightarrow -\frac{4}{3}=B\left(-\frac{1}{3}\right)$
$\displaystyle \Rightarrow B=4$
$\displaystyle \Rightarrow \frac{t}{(t+1)(3t+4)}=-\frac{1}{t+1}+\frac{4}{3t+4}$
$\displaystyle \Rightarrow \frac{x^2}{(x^2+1)(3x^2+4)}=-\frac{1}{x^2+1}+\frac{4}{3x^2+4}$
$\displaystyle \Rightarrow I=-\int\frac{dx}{x^2+1}+4\int\frac{dx}{3x^2+4}$
$\displaystyle =-\tan^{-1}x+4\int\frac{dx}{3\left(x^2+\left(\frac{2}{\sqrt3}\right)^2\right)}$
$\displaystyle =-\tan^{-1}x+\frac{4}{3}\cdot\frac{\sqrt3}{2}\tan^{-1}\left(\frac{\sqrt3 x}{2}\right)+C$
$\displaystyle =-\tan^{-1}x+\frac{2}{\sqrt3}\tan^{-1}\left(\frac{\sqrt3 x}{2}\right)+C$

$\displaystyle \textbf{Question 43: }~\int \frac{3x+5}{x^3-x^2-x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{(3x+5)\,dx}{x^3-x^2-x+1}$
$\displaystyle =\int \frac{(3x+5)\,dx}{x^2(x-1)-1(x-1)}$
$\displaystyle =\int \frac{(3x+5)\,dx}{(x^2-1)(x-1)}$
$\displaystyle =\int \frac{(3x+5)\,dx}{(x-1)^2(x+1)}$
$\displaystyle \text{Let } \frac{3x+5}{(x-1)^2(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$
$\displaystyle \Rightarrow \frac{3x+5}{(x-1)^2(x+1)}=\frac{A(x-1)^2+B(x+1)(x-1)+C(x+1)}{(x-1)^2(x+1)}$
$\displaystyle \Rightarrow 3x+5=A(x^2-2x+1)+B(x^2-1)+Cx+C$
$\displaystyle \Rightarrow 3x+5=(A+B)x^2+(-2A+C)x+(A-B+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0\qquad (1)$
$\displaystyle -2A+C=3\qquad (2)$
$\displaystyle A-B+C=5\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac{1}{2},\; B=-\frac{1}{2},\; C=4$
$\displaystyle \Rightarrow \frac{3x+5}{(x-1)^2(x+1)}=\frac{1}{2(x+1)}-\frac{1}{2(x-1)}+\frac{4}{(x-1)^2}$
$\displaystyle \Rightarrow I=\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{x-1}+4\int (x-1)^{-2}dx$
$\displaystyle =\frac{1}{2}\log|x+1|-\frac{1}{2}\log|x-1|-4\cdot\frac{1}{x-1}+C$
$\displaystyle =\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C$

$\displaystyle \textbf{Question 44: }~\int \frac{x^3-1}{x^3+x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^3-1}{x^3+x}\,dx$
$\displaystyle \text{Degree of numerator equals degree of denominator, so we divide.}$
$\displaystyle \Rightarrow \frac{x^3-1}{x^3+x}=1-\frac{x+1}{x^3+x}$
$\displaystyle \Rightarrow \frac{x^3-1}{x^3+x}=1-\frac{x+1}{x(x^2+1)}\qquad (1)$
$\displaystyle \text{Let } \frac{x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow \frac{x+1}{x(x^2+1)}=\frac{A(x^2+1)+(Bx+C)x}{x(x^2+1)}$
$\displaystyle \Rightarrow x+1=A(x^2+1)+Bx^2+Cx$
$\displaystyle \Rightarrow x+1=(A+B)x^2+Cx+A$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0$
$\displaystyle C=1$
$\displaystyle A=1$
$\displaystyle \Rightarrow B=-1$
$\displaystyle \Rightarrow \frac{x+1}{x(x^2+1)}=\frac{1}{x}+\frac{-x+1}{x^2+1}\qquad (2)$
$\displaystyle \text{Using (1) and (2)}$
$\displaystyle I=\int\left(1-\frac{1}{x}+\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)dx$
$\displaystyle =\int dx-\int\frac{dx}{x}+\int\frac{x\,dx}{x^2+1}-\int\frac{dx}{x^2+1}$
$\displaystyle \text{Put } x^2+1=t$
$\displaystyle \Rightarrow 2x\,dx=dt$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac{1}{2}\log|t|=\frac{1}{2}\log|x^2+1|$
$\displaystyle \Rightarrow I=x-\log|x|+\frac{1}{2}\log|x^2+1|-\tan^{-1}x+C$

$\displaystyle \textbf{Question 45: }~\int \frac{x^2+x+1}{(x+1)^2(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{x^2+x+1}{(x+1)^2(x+2)}\,dx$
$\displaystyle \text{Let } \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$
$\displaystyle \Rightarrow \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^2}{(x+1)^2(x+2)}$
$\displaystyle \Rightarrow x^2+x+1=A(x^2+3x+2)+Bx+2B+C(x^2+2x+1)$
$\displaystyle \Rightarrow x^2+x+1=(A+C)x^2+(3A+B+2C)x+(2A+2B+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+C=1\qquad (1)$
$\displaystyle 3A+B+2C=1\qquad (2)$
$\displaystyle 2A+2B+C=1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=-2,\; B=1,\; C=3$
$\displaystyle \Rightarrow \frac{x^2+x+1}{(x+1)^2(x+2)}=-\frac{2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}$
$\displaystyle \Rightarrow I=-2\int\frac{dx}{x+1}+\int\frac{dx}{(x+1)^2}+3\int\frac{dx}{x+2}$
$\displaystyle =-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+C$

$\displaystyle \textbf{Question 46: }~\int \frac{1}{x(x^4+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{dx}{x(x^4+1)}$
$\displaystyle =\int \frac{x^3\,dx}{x^4(x^4+1)}$
$\displaystyle \text{Put } x^4=t$
$\displaystyle \Rightarrow 4x^3\,dx=dt$
$\displaystyle \Rightarrow x^3\,dx=\frac{dt}{4}$
$\displaystyle \Rightarrow I=\frac{1}{4}\int\frac{dt}{t(t+1)}$
$\displaystyle \text{Let } \frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}$
$\displaystyle \Rightarrow \frac{1}{t(t+1)}=\frac{A(t+1)+Bt}{t(t+1)}$
$\displaystyle \Rightarrow 1=A(t+1)+Bt$
$\displaystyle \text{Putting } t+1=0$
$\displaystyle \Rightarrow t=-1$
$\displaystyle \Rightarrow 1=A\cdot0+B(-1)$
$\displaystyle \Rightarrow B=-1$
$\displaystyle \text{Putting } t=0$
$\displaystyle \Rightarrow 1=A(1)+B\cdot0$
$\displaystyle \Rightarrow A=1$
$\displaystyle \Rightarrow I=\frac{1}{4}\int\frac{dt}{t}-\frac{1}{4}\int\frac{dt}{t+1}$
$\displaystyle =\frac{1}{4}\log|t|-\frac{1}{4}\log|t+1|+C$
$\displaystyle =\frac{1}{4}\log\left|\frac{t}{t+1}\right|+C$
$\displaystyle =\frac{1}{4}\log\left|\frac{x^4}{x^4+1}\right|+C$

$\displaystyle \textbf{Question 47: }~\int \frac{1}{x(x^3+8)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{1}{x(x^3+8)}\,dx$
$\displaystyle \text{We express } \frac{1}{x(x^3+8)}=\frac{A}{x}+\frac{Bx^2+Cx+D}{x^3+8}$
$\displaystyle \Rightarrow 1=A(x^3+8)+(Bx^2+Cx+D)x$
$\displaystyle \Rightarrow 1=Ax^3+8A+Bx^3+Cx^2+Dx$
$\displaystyle \Rightarrow 1=(A+B)x^3+Cx^2+Dx+8A$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=0$
$\displaystyle C=0$
$\displaystyle D=0$
$\displaystyle 8A=1$
$\displaystyle \Rightarrow A=\frac{1}{8},\; B=-\frac{1}{8},\; C=0,\; D=0$
$\displaystyle \Rightarrow \frac{1}{x(x^3+8)}=\frac{1}{8x}-\frac{x^2}{8(x^3+8)}$
$\displaystyle \Rightarrow I=\frac{1}{8}\int\frac{dx}{x}-\frac{1}{8}\int\frac{x^2\,dx}{x^3+8}$
$\displaystyle \text{Put } t=x^3+8$
$\displaystyle \Rightarrow dt=3x^2\,dx$
$\displaystyle \Rightarrow x^2\,dx=\frac{dt}{3}$
$\displaystyle \Rightarrow I=\frac{1}{8}\log|x|-\frac{1}{24}\int\frac{dt}{t}$
$\displaystyle =\frac{1}{8}\log|x|-\frac{1}{24}\log|t|+C$
$\displaystyle =\frac{1}{8}\log|x|-\frac{1}{24}\log|x^3+8|+C$

$\displaystyle \textbf{Question 48: }~\int \frac{3}{(1-x)(1+x^2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{3\,dx}{(1-x)(1+x^2)}$
$\displaystyle =3\int \frac{dx}{(1-x)(1+x^2)}$
$\displaystyle \text{Let } \frac{1}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow \frac{1}{(1-x)(1+x^2)}=\frac{A(x^2+1)+(Bx+C)(1-x)}{(1-x)(1+x^2)}$
$\displaystyle \Rightarrow 1=Ax^2+A+Bx-Bx^2+C-Cx$
$\displaystyle \Rightarrow 1=(A-B)x^2+(B-C)x+(A+C)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A-B=0\qquad (1)$
$\displaystyle B-C=0\qquad (2)$
$\displaystyle A+C=1\qquad (3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle A=\frac12,\; B=\frac12,\; C=\frac12$
$\displaystyle \Rightarrow \frac{1}{(1-x)(1+x^2)}=\frac{1}{2(1-x)}+\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+1)}$
$\displaystyle \Rightarrow I=\frac{3}{2}\int\frac{dx}{1-x}+\frac{3}{2}\int\frac{x\,dx}{x^2+1}+\frac{3}{2}\int\frac{dx}{x^2+1}$
$\displaystyle \text{Put } t=x^2+1$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow \int\frac{x\,dx}{x^2+1}=\frac12\log|t|=\frac12\log|x^2+1|$
$\displaystyle \Rightarrow I=-\frac{3}{2}\log|1-x|+\frac{3}{4}\log|x^2+1|+\frac{3}{2}\tan^{-1}x+C$
$\displaystyle =\frac{3}{4}\left[\log\left|\frac{x^2+1}{(1-x)^2}\right|+2\tan^{-1}x\right]+C$

$\displaystyle \textbf{Question 49: }~\int \frac{\cos x}{(1-\sin x)^3(2+\sin x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } I=\int \frac{\cos x}{(1-\sin x)^3(2+\sin x)}\,dx$
$\displaystyle \text{Let } \sin x=t$
$\displaystyle \Rightarrow \cos x\,dx=dt$
$\displaystyle \Rightarrow I=\int \frac{dt}{(1-t)^3(2+t)}$
$\displaystyle =-\int \frac{dt}{(t-1)^3(t+2)}$
$\displaystyle \text{Let } \frac{1}{(t-1)^3(t+2)}=\frac{A}{t-1}+\frac{B}{(t-1)^2}+\frac{C}{(t-1)^3}+\frac{D}{t+2}\qquad (1)$
$\displaystyle \Rightarrow 1=A(t-1)^2(t+2)+B(t-1)(t+2)+C(t+2)+D(t-1)^3\qquad (2)$
$\displaystyle \text{Putting } t=1 \text{ in (2)}$
$\displaystyle \Rightarrow 1=3C$
$\displaystyle \Rightarrow C=\frac{1}{3}$
$\displaystyle \text{Putting } t=-2 \text{ in (2)}$
$\displaystyle \Rightarrow 1=D(-3)^3$
$\displaystyle \Rightarrow 1=-27D$
$\displaystyle \Rightarrow D=-\frac{1}{27}$
$\displaystyle \text{Putting } t=0 \text{ in (2)}$
$\displaystyle \Rightarrow 1=2A-2B+2C-D$
$\displaystyle \Rightarrow 1=2A-2B+\frac{2}{3}+\frac{1}{27}$
$\displaystyle \Rightarrow 2A-2B=\frac{8}{27}$
$\displaystyle \Rightarrow A-B=\frac{4}{27}\qquad (3)$
$\displaystyle \text{Putting } t=2 \text{ in (2)}$
$\displaystyle \Rightarrow 1=4A+4B+4C+D$
$\displaystyle \Rightarrow 1=4A+4B+\frac{4}{3}-\frac{1}{27}$
$\displaystyle \Rightarrow A+B=-\frac{2}{27}\qquad (4)$
$\displaystyle \text{Solving (3) and (4), we get}$
$\displaystyle A=\frac{1}{27},\; B=-\frac{1}{9}$
$\displaystyle \text{Substituting } A,B,C,D \text{ in (1)}$
$\displaystyle \Rightarrow \frac{1}{(t-1)^3(t+2)}=\frac{1}{27(t-1)}-\frac{1}{9(t-1)^2}+\frac{1}{3(t-1)^3}-\frac{1}{27(t+2)}$
$\displaystyle \Rightarrow I=-\int\left[\frac{1}{27(t-1)}-\frac{1}{9(t-1)^2}+\frac{1}{3(t-1)^3}-\frac{1}{27(t+2)}\right]dt$
$\displaystyle =-\left[\frac{1}{27}\log|t-1|+\frac{1}{9(t-1)}-\frac{1}{6(t-1)^2}-\frac{1}{27}\log|t+2|\right]+C$
$\displaystyle =-\frac{1}{27}\log|\sin x-1|+\frac{1}{9(1-\sin x)}+\frac{1}{6(1-\sin x)^2}+\frac{1}{27}\log|2+\sin x|+C$

$\displaystyle \textbf{Question 50: }~\int \frac{2x^2+1}{x^2(x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{2x^2+1}{x^2(x^2+4)}\,dx$
$\displaystyle \text{We express } \frac{2x^2+1}{x^2(x^2+4)}=\frac{A}{x^2}+\frac{B}{x^2+4}$
$\displaystyle \Rightarrow 2x^2+1=A(x^2+4)+Bx^2$
$\displaystyle \Rightarrow 2x^2+1=(A+B)x^2+4A$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=2$
$\displaystyle 4A=1$
$\displaystyle \Rightarrow A=\frac14,\; B=\frac74$
$\displaystyle \Rightarrow \frac{2x^2+1}{x^2(x^2+4)}=\frac{1}{4x^2}+\frac{7}{4(x^2+4)}$
$\displaystyle \Rightarrow I=\frac14\int\frac{dx}{x^2}+\frac74\int\frac{dx}{x^2+4}$
$\displaystyle =\frac14\int x^{-2}\,dx+\frac74\int\frac{dx}{x^2+2^2}$
$\displaystyle =-\frac{1}{4x}+\frac74\cdot\frac12\tan^{-1}\left(\frac{x}{2}\right)+C$
$\displaystyle =-\frac{1}{4x}+\frac78\tan^{-1}\left(\frac{x}{2}\right)+C$

$\displaystyle \textbf{Question 51: }~\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{\cos x\,dx}{(1-\sin x)(2-\sin x)}$
$\displaystyle \text{Let }\sin x=t$
$\displaystyle \cos x\,dx=dt$
$\displaystyle I=\int\frac{dt}{(1-t)(2-t)}$
$\displaystyle I=\int\frac{dt}{(t-1)(t-2)}$
$\displaystyle \frac{1}{(t-1)(t-2)}=\frac{A}{t-1}+\frac{B}{t-2}$
$\displaystyle \frac{1}{(t-1)(t-2)}=\frac{A(t-2)+B(t-1)}{(t-1)(t-2)}$
$\displaystyle 1=A(t-2)+B(t-1)$
$\displaystyle \text{Putting }t=1$
$\displaystyle 1=A(1-2)$
$\displaystyle A=-1$
$\displaystyle \text{Putting }t=2$
$\displaystyle 1=B(2-1)$
$\displaystyle B=1$
$\displaystyle I=\int\left(-\frac{1}{t-1}+\frac{1}{t-2}\right)dt$
$\displaystyle I=-\log|t-1|+\log|t-2|+C$
$\displaystyle I=\log\left|\frac{t-2}{t-1}\right|+C$
$\displaystyle I=\log\left|\frac{\sin x-2}{\sin x-1}\right|+C$
$\displaystyle I=\log\left|\frac{2-\sin x}{1-\sin x}\right|+C$

$\displaystyle \textbf{Question 52: }~\int \frac{2x+1}{(x-2)(x-3)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{2x+1}{(x-2)(x-3)}\,dx$
$\displaystyle \text{Let }\frac{2x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
$\displaystyle \frac{2x+1}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}$
$\displaystyle 2x+1=A(x-3)+B(x-2)$
$\displaystyle \text{Putting }x=3$
$\displaystyle 7=B(3-2)$
$\displaystyle B=7$
$\displaystyle \text{Putting }x=2$
$\displaystyle 5=A(2-3)$
$\displaystyle A=-5$
$\displaystyle I=\int\left(-\frac{5}{x-2}+\frac{7}{x-3}\right)dx$
$\displaystyle I=-5\log|x-2|+7\log|x-3|+C$
$\displaystyle I=\log|x-3|^{7}-\log|x-2|^{5}+C$
$\displaystyle I=\log\left|\frac{(x-3)^7}{(x-2)^5}\right|+C$

$\displaystyle \textbf{Question 53: }~\int \frac{1}{(x^2+1)(x^2+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{(x^2+1)(x^2+2)}$
$\displaystyle \text{Let }x^2=t$
$\displaystyle \frac{1}{(x^2+1)(x^2+2)}=\frac{1}{(t+1)(t+2)}$
$\displaystyle \text{Let }\frac{1}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}$
$\displaystyle \frac{1}{(t+1)(t+2)}=\frac{A(t+2)+B(t+1)}{(t+1)(t+2)}$
$\displaystyle 1=A(t+2)+B(t+1)$
$\displaystyle \text{Putting }t=-2$
$\displaystyle 1=B(-1)$
$\displaystyle B=-1$
$\displaystyle \text{Putting }t=-1$
$\displaystyle 1=A(1)$
$\displaystyle A=1$
$\displaystyle \frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}$
$\displaystyle \frac{1}{(x^2+1)(x^2+2)}=\frac{1}{x^2+1}-\frac{1}{x^2+2}$
$\displaystyle I=\int\frac{dx}{x^2+1}-\int\frac{dx}{x^2+2}$
$\displaystyle I=\tan^{-1}x-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+C$

$\displaystyle \textbf{Question 54: }~\int \frac{1}{x(x^4-1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{x(x^4-1)}$
$\displaystyle I=\int\frac{x^3\,dx}{x^4(x^4-1)}$
$\displaystyle \text{Putting }x^4=t$
$\displaystyle 4x^3\,dx=dt$
$\displaystyle x^3\,dx=\frac{dt}{4}$
$\displaystyle I=\frac{1}{4}\int\frac{dt}{t(t-1)}$
$\displaystyle \text{Let }\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$
$\displaystyle \frac{1}{t(t-1)}=\frac{A(t-1)+Bt}{t(t-1)}$
$\displaystyle 1=A(t-1)+Bt$
$\displaystyle \text{Putting }t=1$
$\displaystyle 1=B$
$\displaystyle B=1$
$\displaystyle \text{Putting }t=0$
$\displaystyle 1=-A$
$\displaystyle A=-1$
$\displaystyle I=\frac{1}{4}\int\left(-\frac{1}{t}+\frac{1}{t-1}\right)dt$
$\displaystyle I=-\frac{1}{4}\log|t|+\frac{1}{4}\log|t-1|+C$
$\displaystyle I=\frac{1}{4}\log\left|\frac{t-1}{t}\right|+C$
$\displaystyle I=\frac{1}{4}\log\left|\frac{x^4-1}{x^4}\right|+C$

$\displaystyle \textbf{Question 55: }~\int \frac{1}{4^x-1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{x^4-1}$
$\displaystyle I=\int\frac{dx}{(x^2-1)(x^2+1)}$
$\displaystyle I=\int\frac{dx}{(x-1)(x+1)(x^2+1)}$
$\displaystyle \text{Let }\frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$
$\displaystyle \frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A(x^2+1)(x+1)+B(x^2+1)(x-1)+(Cx+D)(x^2-1)}{(x-1)(x+1)(x^2+1)}$
$\displaystyle 1=A(x^2+1)(x+1)+B(x^2+1)(x-1)+(Cx+D)(x^2-1)$
$\displaystyle 1=A(x^3+x^2+x+1)+B(x^3-x^2+x-1)+(Cx^3-Cx+Dx^2-D)$
$\displaystyle 1=(A+B+C)x^3+(A-B+D)x^2+(A+B-C)x+(A-B-D)$
$\displaystyle \text{Equating coefficients of like powers of }x$
$\displaystyle A+B+C=0$
$\displaystyle A-B+D=0$
$\displaystyle A+B-C=0$
$\displaystyle A-B-D=1$
$\displaystyle \text{Solving, we get }A=\frac{1}{4},\;B=-\frac{1}{4},\;C=0,\;D=-\frac{1}{2}$
$\displaystyle \frac{1}{(x-1)(x+1)(x^2+1)}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2(x^2+1)}$
$\displaystyle I=\frac{1}{4}\int\frac{dx}{x-1}-\frac{1}{4}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{x^2+1}$
$\displaystyle I=\frac{1}{4}\log|x-1|-\frac{1}{4}\log|x+1|-\frac{1}{2}\tan^{-1}x+C$
$\displaystyle I=\frac{1}{4}\log\left|\frac{x-1}{x+1}\right|-\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 56: }~\int \frac{2x}{(x^2+1)(x^2+2)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{2x}{(x^2+1)(x^2+2)^2}\,dx$
$\displaystyle \text{Let }x^2=y$
$\displaystyle 2x\,dx=dy$
$\displaystyle I=\int\frac{dy}{(y+1)(y+2)^2}$
$\displaystyle \text{Let }\frac{1}{(y+1)(y+2)^2}=\frac{A}{y+1}+\frac{B}{y+2}+\frac{C}{(y+2)^2}$
$\displaystyle \frac{1}{(y+1)(y+2)^2}=\frac{A(y+2)^2+B(y+1)(y+2)+C(y+1)}{(y+1)(y+2)^2}$
$\displaystyle 1=A(y+2)^2+B(y+1)(y+2)+C(y+1)$
$\displaystyle \text{Putting }y=-2$
$\displaystyle 1=C(-1)$
$\displaystyle C=-1$
$\displaystyle \text{Putting }y=-1$
$\displaystyle 1=A(1)$
$\displaystyle A=1$
$\displaystyle \text{Putting }y=0$
$\displaystyle 1=4A+2B+C$
$\displaystyle 1=4+2B-1$
$\displaystyle 1=3+2B$
$\displaystyle B=-1$
$\displaystyle \frac{1}{(y+1)(y+2)^2}=\frac{1}{y+1}-\frac{1}{y+2}-\frac{1}{(y+2)^2}$
$\displaystyle I=\int\frac{dy}{y+1}-\int\frac{dy}{y+2}-\int\frac{dy}{(y+2)^2}$
$\displaystyle I=\log|y+1|-\log|y+2|+\frac{1}{y+2}+C$
$\displaystyle I=\log|x^2+1|-\log|x^2+2|+\frac{1}{x^2+2}+C$

$\displaystyle \textbf{Question 57: }~\int \frac{x^2}{(x-1)(x^2+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{x^2}{(x-1)(x^2+1)}\,dx$
$\displaystyle \text{Let }\frac{x^2}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$
$\displaystyle \frac{x^2}{(x-1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$
$\displaystyle x^2=A(x^2+1)+(Bx+C)(x-1)$
$\displaystyle x^2=(A+B)x^2+(C-B)x+(A-C)$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle A+B=1$
$\displaystyle C-B=0$
$\displaystyle A-C=0$
$\displaystyle \text{Solving, we get }A=\frac{1}{2},\;B=\frac{1}{2},\;C=\frac{1}{2}$
$\displaystyle I=\frac{1}{2}\int\frac{dx}{x-1}+\frac{1}{2}\int\frac{x\,dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x^2+1}$
$\displaystyle I=\frac{1}{2}\log|x-1|+\frac{1}{4}\log(x^2+1)+\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 58: }~\int \frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int\frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx$
$\displaystyle \frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{A}{x^2+a^2}+\frac{B}{x^2+b^2}$
$\displaystyle x^2=A(x^2+b^2)+B(x^2+a^2)$
$\displaystyle x^2=(A+B)x^2+(Ab^2+Ba^2)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=1$
$\displaystyle Ab^2+Ba^2=0$
$\displaystyle A=\frac{a^2}{a^2-b^2}$
$\displaystyle B=\frac{b^2}{b^2-a^2}$
$\displaystyle I=\int\left(\frac{a^2}{(a^2-b^2)(x^2+a^2)}+\frac{b^2}{(b^2-a^2)(x^2+b^2)}\right)dx$
$\displaystyle I=\frac{a^2}{a^2-b^2}\int\frac{dx}{x^2+a^2}-\frac{b^2}{a^2-b^2}\int\frac{dx}{x^2+b^2}$
$\displaystyle I=\frac{a}{a^2-b^2}\tan^{-1}\left(\frac{x}{a}\right)-\frac{b}{a^2-b^2}\tan^{-1}\left(\frac{x}{b}\right)+C$

$\displaystyle \textbf{Question 59: }~\int \frac{1}{\cos x(5-4\sin x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{\cos x(5-4\sin x)}$
$\displaystyle I=\int\frac{\cos x\,dx}{\cos^2 x(5-4\sin x)}$
$\displaystyle I=\int\frac{\cos x\,dx}{(1-\sin^2 x)(5-4\sin x)}$
$\displaystyle I=\int\frac{\cos x\,dx}{(1-\sin x)(1+\sin x)(5-4\sin x)}$
$\displaystyle \text{Putting }\sin x=t$
$\displaystyle \cos x\,dx=dt$
$\displaystyle I=\int\frac{dt}{(1-t)(1+t)(5-4t)}$
$\displaystyle \text{Let }\frac{1}{(1-t)(1+t)(5-4t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{5-4t}$
$\displaystyle \frac{1}{(1-t)(1+t)(5-4t)}=\frac{A(1+t)(5-4t)+B(1-t)(5-4t)+C(1-t)(1+t)}{(1-t)(1+t)(5-4t)}$
$\displaystyle 1=A(1+t)(5-4t)+B(1-t)(5-4t)+C(1-t)(1+t)$
$\displaystyle \text{Putting }t=-1$
$\displaystyle 1=B(2)(9)$
$\displaystyle B=\frac{1}{18}$
$\displaystyle \text{Putting }t=1$
$\displaystyle 1=A(2)(1)$
$\displaystyle A=\frac{1}{2}$
$\displaystyle \text{Putting }5-4t=0$
$\displaystyle t=\frac{5}{4}$
$\displaystyle 1=C\left(1-\frac{5}{4}\right)\left(1+\frac{5}{4}\right)$
$\displaystyle 1=C\left(-\frac{1}{4}\right)\left(\frac{9}{4}\right)$
$\displaystyle C=-\frac{16}{9}$
$\displaystyle I=\frac{1}{2}\int\frac{dt}{1-t}+\frac{1}{18}\int\frac{dt}{1+t}-\frac{16}{9}\int\frac{dt}{5-4t}$
$\displaystyle I=-\frac{1}{2}\log|1-t|+\frac{1}{18}\log|1+t|+\frac{4}{9}\log|5-4t|+C$
$\displaystyle I=\frac{1}{18}\log|1+\sin x|-\frac{1}{2}\log|1-\sin x|+\frac{4}{9}\log|5-4\sin x|+C$

$\displaystyle \textbf{Question 60: }~\int \frac{1}{\sin x(3+2\cos x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{\sin x(3+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{\sin^2 x(3+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{(1-\cos^2 x)(3+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(3+2\cos x)}$
$\displaystyle \text{Putting }\cos x=t$
$\displaystyle -\sin x\,dx=dt$
$\displaystyle \sin x\,dx=-dt$
$\displaystyle I=\int\frac{-dt}{(1-t)(1+t)(3+2t)}$
$\displaystyle I=\int\frac{dt}{(t-1)(t+1)(3+2t)}$
$\displaystyle \text{Let }\frac{1}{(t-1)(t+1)(3+2t)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{3+2t}$
$\displaystyle \frac{1}{(t-1)(t+1)(3+2t)}=\frac{A(t+1)(3+2t)+B(t-1)(3+2t)+C(t+1)(t-1)}{(t-1)(t+1)(3+2t)}$
$\displaystyle 1=A(t+1)(3+2t)+B(t-1)(3+2t)+C(t+1)(t-1)$
$\displaystyle \text{Putting }t=-1$
$\displaystyle 1=B(-2)(1)$
$\displaystyle B=-\frac{1}{2}$
$\displaystyle \text{Putting }t=1$
$\displaystyle 1=A(2)(5)$
$\displaystyle A=\frac{1}{10}$
$\displaystyle \text{Putting }3+2t=0$
$\displaystyle t=-\frac{3}{2}$
$\displaystyle 1=C\left(-\frac{1}{2}\right)\left(-\frac{5}{2}\right)$
$\displaystyle C=\frac{4}{5}$
$\displaystyle I=\frac{1}{10}\int\frac{dt}{t-1}-\frac{1}{2}\int\frac{dt}{t+1}+\frac{4}{5}\int\frac{dt}{3+2t}$
$\displaystyle I=\frac{1}{10}\log|t-1|-\frac{1}{2}\log|t+1|+\frac{2}{5}\log|3+2t|+C$
$\displaystyle I=\frac{1}{10}\log|\cos x-1|-\frac{1}{2}\log|\cos x+1|+\frac{2}{5}\log|3+2\cos x|+C$

$\displaystyle \textbf{Question 61: }~\int \frac{1}{\sin x+\sin 2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{dx}{\sin x+\sin 2x}$
$\displaystyle I=\int\frac{dx}{\sin x+2\sin x\cos x}$
$\displaystyle I=\int\frac{dx}{\sin x(1+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{\sin^2 x(1+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{(1-\cos^2 x)(1+2\cos x)}$
$\displaystyle I=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$
$\displaystyle \text{Putting }\cos x=t$
$\displaystyle -\sin x\,dx=dt$
$\displaystyle \sin x\,dx=-dt$
$\displaystyle I=\int\frac{-dt}{(1-t)(1+t)(1+2t)}$
$\displaystyle I=\int\frac{dt}{(t-1)(t+1)(1+2t)}$
$\displaystyle \text{Let }\frac{1}{(t-1)(t+1)(1+2t)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{1+2t}$
$\displaystyle \frac{1}{(t-1)(t+1)(1+2t)}=\frac{A(t+1)(1+2t)+B(t-1)(1+2t)+C(t-1)(t+1)}{(t-1)(t+1)(1+2t)}$
$\displaystyle 1=A(t+1)(1+2t)+B(t-1)(1+2t)+C(t-1)(t+1)$
$\displaystyle \text{Putting }t=-1$
$\displaystyle 1=B(-2)(-1)$
$\displaystyle B=\frac{1}{2}$
$\displaystyle \text{Putting }t=1$
$\displaystyle 1=A(2)(3)$
$\displaystyle A=\frac{1}{6}$
$\displaystyle \text{Putting }1+2t=0$
$\displaystyle t=-\frac{1}{2}$
$\displaystyle 1=C\left(-\frac{3}{2}\right)\left(\frac{1}{2}\right)$
$\displaystyle C=-\frac{4}{3}$
$\displaystyle I=\frac{1}{6}\int\frac{dt}{t-1}+\frac{1}{2}\int\frac{dt}{t+1}-\frac{4}{3}\int\frac{dt}{1+2t}$
$\displaystyle I=\frac{1}{6}\log|t-1|+\frac{1}{2}\log|t+1|-\frac{2}{3}\log|1+2t|+C$
$\displaystyle I=\frac{1}{6}\log|\cos x-1|+\frac{1}{2}\log|\cos x+1|-\frac{2}{3}\log|1+2\cos x|+C$

$\displaystyle \textbf{Question 62: }~\int \frac{x+1}{x(1+x e^x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{x+1}{x(1+x\mathrm{e}^x)}\,dx$
$\displaystyle I=\int\frac{\mathrm{e}^x(x+1)}{\mathrm{e}^x x(1+x\mathrm{e}^x)}\,dx$
$\displaystyle \text{Put }\mathrm{e}^x=t$
$\displaystyle \mathrm{e}^x(x+1)\,dx=dt$
$\displaystyle I=\int\frac{dt}{t(1+t)}$
$\displaystyle \text{Let }\frac{1}{t(1+t)}=\frac{A}{t}+\frac{B}{1+t}$
$\displaystyle \frac{1}{t(1+t)}=\frac{A(1+t)+Bt}{t(1+t)}$
$\displaystyle 1=A(1+t)+Bt$
$\displaystyle \text{Putting }t=0$
$\displaystyle 1=A$
$\displaystyle A=1$
$\displaystyle \text{Putting }t=-1$
$\displaystyle 1=-B$
$\displaystyle B=-1$
$\displaystyle I=\int\left(\frac{1}{t}-\frac{1}{1+t}\right)dt$
$\displaystyle I=\log|t|-\log|1+t|+C$
$\displaystyle I=\log\left|\frac{t}{1+t}\right|+C$
$\displaystyle I=\log\left|\frac{x\mathrm{e}^x}{1+x\mathrm{e}^x}\right|+C$

$\displaystyle \textbf{Question 63: }~\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx$
$\displaystyle \text{Put }x^2=t$
$\displaystyle \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}$
$\displaystyle \frac{t^2+3t+2}{t^2+7t+12}$
$\displaystyle \text{Since the degrees of numerator and denominator are equal, divide numerator by denominator}$
$\displaystyle \frac{t^2+3t+2}{t^2+7t+12}=1-\frac{4t+10}{t^2+7t+12}$
$\displaystyle \frac{t^2+3t+2}{t^2+7t+12}=1-\frac{4t+10}{(t+3)(t+4)}$
$\displaystyle \text{Let }\frac{4t+10}{(t+3)(t+4)}=\frac{A}{t+3}+\frac{B}{t+4}$
$\displaystyle \frac{4t+10}{(t+3)(t+4)}=\frac{A(t+4)+B(t+3)}{(t+3)(t+4)}$
$\displaystyle 4t+10=A(t+4)+B(t+3)$
$\displaystyle \text{Putting }t=-4$
$\displaystyle -16+10=B(-1)$
$\displaystyle B=6$
$\displaystyle \text{Putting }t=-3$
$\displaystyle -12+10=A(1)$
$\displaystyle A=-2$
$\displaystyle \frac{4t+10}{(t+3)(t+4)}=-\frac{2}{t+3}+\frac{6}{t+4}$
$\displaystyle \frac{t^2+3t+2}{t^2+7t+12}=1+\frac{2}{t+3}-\frac{6}{t+4}$
$\displaystyle I=\int dx+2\int\frac{dx}{x^2+3}-6\int\frac{dx}{x^2+4}$
$\displaystyle I=x+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)-3\tan^{-1}\left(\frac{x}{2}\right)+C$

$\displaystyle \textbf{Question 64: }~\int \frac{4x^4+3}{(x^2+2)(x^2+3)(x^2+4)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{4x^4+3}{(x^2+2)(x^2+3)(x^2+4)}\,dx$
$\displaystyle \text{Put }x^2=t$
$\displaystyle \frac{4x^4+3}{(x^2+2)(x^2+3)(x^2+4)}=\frac{4t^2+3}{(t+2)(t+3)(t+4)}$
$\displaystyle \text{Let }\frac{4t^2+3}{(t+2)(t+3)(t+4)}=\frac{A}{t+2}+\frac{B}{t+3}+\frac{C}{t+4}$
$\displaystyle \frac{4t^2+3}{(t+2)(t+3)(t+4)}=\frac{A(t+3)(t+4)+B(t+2)(t+4)+C(t+2)(t+3)}{(t+2)(t+3)(t+4)}$
$\displaystyle 4t^2+3=A(t+3)(t+4)+B(t+2)(t+4)+C(t+2)(t+3)$
$\displaystyle \text{Putting }t=-3$
$\displaystyle 4(-3)^2+3=B(-1)(1)$
$\displaystyle B=-39$
$\displaystyle \text{Putting }t=-2$
$\displaystyle 4(-2)^2+3=A(1)(2)$
$\displaystyle A=\frac{19}{2}$
$\displaystyle \text{Putting }t=-4$
$\displaystyle 4(-4)^2+3=C(-2)(-1)$
$\displaystyle C=\frac{67}{2}$
$\displaystyle \frac{4t^2+3}{(t+2)(t+3)(t+4)}=\frac{19}{2(t+2)}-\frac{39}{t+3}+\frac{67}{2(t+4)}$
$\displaystyle I=\frac{19}{2}\int\frac{dx}{x^2+2}-39\int\frac{dx}{x^2+3}+\frac{67}{2}\int\frac{dx}{x^2+4}$
$\displaystyle I=\frac{19}{2\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{39}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{67}{4}\tan^{-1}\left(\frac{x}{2}\right)+C$

$\displaystyle \textbf{Question 65: }~\int \frac{x^4}{(x-1)(x^2+1)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{x^4}{(x-1)(x^2+1)}\,dx$
$\displaystyle I=\int\frac{x^4-1+1}{(x-1)(x^2+1)}\,dx$
$\displaystyle I=\int\frac{x^4-1}{(x-1)(x^2+1)}\,dx+\int\frac{dx}{(x-1)(x^2+1)}$
$\displaystyle I=\int\frac{(x^2-1)(x^2+1)}{(x-1)(x^2+1)}\,dx+\int\frac{dx}{(x-1)(x^2+1)}$
$\displaystyle I=\int\frac{(x-1)(x+1)}{(x-1)}\,dx+\int\frac{dx}{(x-1)(x^2+1)}$
$\displaystyle I=\int(x+1)\,dx+\int\frac{dx}{(x-1)(x^2+1)}$
$\displaystyle \text{Let }\frac{1}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$
$\displaystyle \frac{1}{(x-1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$
$\displaystyle 1=A(x^2+1)+(Bx+C)(x-1)$
$\displaystyle 1=(A+B)x^2+(C-B)x+(A-C)$
$\displaystyle \text{Equating coefficients of like powers of }x$
$\displaystyle A+B=0$
$\displaystyle C-B=0$
$\displaystyle A-C=1$
$\displaystyle \text{Solving, we get }A=\frac{1}{2},\;B=-\frac{1}{2},\;C=-\frac{1}{2}$
$\displaystyle \frac{1}{(x-1)(x^2+1)}=\frac{1}{2(x-1)}-\frac{1}{2}\frac{x}{x^2+1}-\frac{1}{2}\frac{1}{x^2+1}$
$\displaystyle I=\int(x+1)\,dx+\frac{1}{2}\int\frac{dx}{x-1}-\frac{1}{2}\int\frac{x\,dx}{x^2+1}-\frac{1}{2}\int\frac{dx}{x^2+1}$
$\displaystyle \text{Putting }x^2+1=t$
$\displaystyle 2x\,dx=dt$
$\displaystyle x\,dx=\frac{dt}{2}$
$\displaystyle I=\frac{x^2}{2}+x+\frac{1}{2}\log|x-1|-\frac{1}{4}\log|t|-\frac{1}{2}\tan^{-1}x+C$
$\displaystyle I=\frac{x^2}{2}+x+\frac{1}{2}\log|x-1|-\frac{1}{4}\log(x^2+1)-\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 66: }~\int \frac{x^2}{x^4-x^2-12}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int\frac{x^2}{x^4-x^2-12}\,dx$
$\displaystyle \frac{x^4-x^2-12}{ }=(x^2-4)(x^2+3)$
$\displaystyle \frac{x^2}{x^4-x^2-12}=\frac{x^2}{(x^2-4)(x^2+3)}$
$\displaystyle \text{Let }\frac{x^2}{(x^2-4)(x^2+3)}=\frac{A}{x^2-4}+\frac{B}{x^2+3}$
$\displaystyle \frac{x^2}{(x^2-4)(x^2+3)}=\frac{A(x^2+3)+B(x^2-4)}{(x^2-4)(x^2+3)}$
$\displaystyle x^2=A(x^2+3)+B(x^2-4)$
$\displaystyle x^2=(A+B)x^2+(3A-4B)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=1$
$\displaystyle 3A-4B=0$
$\displaystyle A=\frac{4}{7}$
$\displaystyle B=\frac{3}{7}$
$\displaystyle I=\int\left(\frac{4}{7}\frac{1}{x^2-4}+\frac{3}{7}\frac{1}{x^2+3}\right)dx$
$\displaystyle I=\frac{4}{7}\int\frac{dx}{x^2-4}+\frac{3}{7}\int\frac{dx}{x^2+3}$
$\displaystyle I=\frac{1}{7}\log\left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$

$\displaystyle \textbf{Question 67: }~\int \frac{x^2}{1-x^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int\frac{x^2}{1-x^4}\,dx$
$\displaystyle \frac{x^2}{1-x^4}=\frac{x^2}{(1-x^2)(1+x^2)}$
$\displaystyle \text{Let }\frac{x^2}{(1-x^2)(1+x^2)}=\frac{A}{1-x^2}+\frac{B}{1+x^2}$
$\displaystyle \frac{x^2}{(1-x^2)(1+x^2)}=\frac{A(1+x^2)+B(1-x^2)}{(1-x^2)(1+x^2)}$
$\displaystyle x^2=A(1+x^2)+B(1-x^2)$
$\displaystyle x^2=(A-B)x^2+(A+B)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A-B=1$
$\displaystyle A+B=0$
$\displaystyle A=\frac{1}{2}$
$\displaystyle B=-\frac{1}{2}$
$\displaystyle I=\int\left(\frac{1}{2}\frac{1}{1-x^2}-\frac{1}{2}\frac{1}{1+x^2}\right)dx$
$\displaystyle I=\frac{1}{2}\int\frac{dx}{1-x^2}-\frac{1}{2}\int\frac{dx}{1+x^2}$
$\displaystyle I=\frac{1}{4}\log\left|\frac{1+x}{1-x}\right|-\frac{1}{2}\tan^{-1}x+C$

$\displaystyle \textbf{Question 68: }~\int \frac{x^2}{x^4+x^2-2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int\frac{x^2}{x^4+x^2-2}\,dx$
$\displaystyle \frac{x^4+x^2-2}{ }=(x^2+2)(x^2-1)$
$\displaystyle \frac{x^2}{x^4+x^2-2}=\frac{x^2}{(x^2+2)(x^2-1)}$
$\displaystyle \text{Let }\frac{x^2}{(x^2+2)(x^2-1)}=\frac{A}{x^2+2}+\frac{B}{x^2-1}$
$\displaystyle \frac{x^2}{(x^2+2)(x^2-1)}=\frac{A(x^2-1)+B(x^2+2)}{(x^2+2)(x^2-1)}$
$\displaystyle x^2=A(x^2-1)+B(x^2+2)$
$\displaystyle x^2=(A+B)x^2+(-A+2B)$
$\displaystyle \text{Equating coefficients of like terms}$
$\displaystyle A+B=1$
$\displaystyle -A+2B=0$
$\displaystyle A=\frac{2}{3}$
$\displaystyle B=\frac{1}{3}$
$\displaystyle I=\int\left(\frac{2}{3}\frac{1}{x^2+2}+\frac{1}{3}\frac{1}{x^2-1}\right)dx$
$\displaystyle I=\frac{2}{3}\int\frac{dx}{x^2+2}+\frac{1}{3}\int\frac{dx}{x^2-1}$
$\displaystyle I=\frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{6}\log\left|\frac{x-1}{x+1}\right|+C$

$\displaystyle \textbf{Question 69: }~\int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\,dx$
$\displaystyle \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}=\frac{(x^2+3-2)(x^2-5+9)}{(x^2+3)(x^2-5)}$
$\displaystyle \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}=\frac{(x^2+3)(x^2-5)+9(x^2+3)-2(x^2-5)-18}{(x^2+3)(x^2-5)}$
$\displaystyle \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}=1+\frac{9}{x^2-5}-\frac{2}{x^2+3}-\frac{18}{(x^2+3)(x^2-5)}$
$\displaystyle \text{Let }I_1=\int\frac{dx}{(x^2+3)(x^2-5)}$
$\displaystyle \text{Put }x^2=y$
$\displaystyle \frac{1}{(y+3)(y-5)}=\frac{A}{y+3}+\frac{B}{y-5}$
$\displaystyle \frac{1}{(y+3)(y-5)}=\frac{A(y-5)+B(y+3)}{(y+3)(y-5)}$
$\displaystyle 1=(A+B)y-(5A-3B)$
$\displaystyle \text{Comparing coefficients}$
$\displaystyle A+B=0$
$\displaystyle 5A-3B=1$
$\displaystyle A=-\frac{1}{8}$
$\displaystyle B=\frac{1}{8}$
$\displaystyle I_1=\int\left(-\frac{1}{8}\frac{1}{x^2+3}+\frac{1}{8}\frac{1}{x^2-5}\right)dx$
$\displaystyle I=\int\left[1+\frac{9}{x^2-5}-\frac{2}{x^2+3}-18\left(-\frac{1}{8}\frac{1}{x^2+3}+\frac{1}{8}\frac{1}{x^2-5}\right)\right]dx$
$\displaystyle I=\int\left[1+\frac{27}{4(x^2-5)}+\frac{1}{x^2+3}\right]dx$
$\displaystyle I=\int dx+\frac{27}{4}\int\frac{dx}{x^2-5}+\int\frac{dx}{x^2+3}$
$\displaystyle I=x+\frac{27}{8\sqrt{5}}\log\left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$