Class 12: Indefinite Integrals

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

$\displaystyle \textbf{Question 1: }~\int \frac{x}{x^2+3x+2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x}{x^2+3x+2}\,dx$
$\displaystyle x=A\frac{d}{dx}(x^2+3x+2)+B$
$\displaystyle x=A(2x+3)+B$
$\displaystyle x=2Ax+3A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x\text{, we get}$
$\displaystyle 2A=1 \Rightarrow A=\frac12$
$\displaystyle 3A+B=0$
$\displaystyle \frac32+B=0$
$\displaystyle B=-\frac32$
$\displaystyle x=\frac12(2x+3)-\frac32$
$\displaystyle \text{Now, } \int \frac{x}{x^2+3x+2}\,dx$
$\displaystyle =\int \left[\frac{\frac12(2x+3)-\frac32}{x^2+3x+2}\right]dx$
$\displaystyle =\frac12\int \frac{2x+3}{x^2+3x+2}\,dx-\frac32\int \frac{dx}{x^2+3x+2}$
$\displaystyle =\frac12\int \frac{2x+3}{x^2+3x+2}\,dx-\frac32\int \frac{dx}{\left(x+\frac32\right)^2-\left(\frac32\right)^2+2}$
$\displaystyle =\frac12\int \frac{2x+3}{x^2+3x+2}\,dx-\frac32\int \frac{dx}{\left(x+\frac32\right)^2-\frac14}$
$\displaystyle =\frac12\log|x^2+3x+2|-\frac32\cdot\frac12\log\left|\frac{x+\frac32-\frac12}{x+\frac32+\frac12}\right|+C$
$\displaystyle =\frac12\log|x^2+3x+2|-\frac32\log\left|\frac{x+1}{x+2}\right|+C$
$\\$
$\displaystyle \textbf{Question 2: }~\int \frac{x+1}{x^2+x+3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x+1}{x^2+x+3}\,dx$
$\displaystyle x+1=A\frac{d}{dx}(x^2+x+3)+B$
$\displaystyle x+1=A(2x+1)+B$
$\displaystyle x+1=2Ax+A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 2A=1$
$\displaystyle A=\frac12$
$\displaystyle A+B=1$
$\displaystyle \frac12+B=1$
$\displaystyle B=\frac12$
$\displaystyle x+1=\frac12(2x+1)+\frac12$
$\displaystyle \text{Now, } \int \frac{x+1}{x^2+x+3}\,dx$
$\displaystyle =\int\left[\frac{\frac12(2x+1)}{x^2+x+3}+\frac12\frac{1}{x^2+x+3}\right]dx$
$\displaystyle =\frac12\int\frac{2x+1}{x^2+x+3}\,dx+\frac12\int\frac{dx}{x^2+x+3}$
$\displaystyle =\frac12\int\frac{2x+1}{x^2+x+3}\,dx+\frac12\int\frac{dx}{\left(x+\frac12\right)^2-\left(\frac12\right)^2+3}$
$\displaystyle =\frac12\int\frac{2x+1}{x^2+x+3}\,dx+\frac12\int\frac{dx}{\left(x+\frac12\right)^2+\frac{11}{4}}$
$\displaystyle =\frac12\log|x^2+x+3|+\frac12\cdot\frac{2}{\sqrt{11}}\tan^{-1}\left(\frac{x+\frac12}{\frac{\sqrt{11}}{2}}\right)+C$
$\displaystyle =\frac12\log|x^2+x+3|+\frac{1}{\sqrt{11}}\tan^{-1}\left(\frac{2x+1}{\sqrt{11}}\right)+C$
$\\$
$\displaystyle \textbf{Question 3: }~\int \frac{x-3}{x^2+2x-4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x-3}{x^2+2x-4}\,dx$
$\displaystyle x-3=A\frac{d}{dx}(x^2+2x-4)+B$
$\displaystyle x-3=A(2x+2)+B$
$\displaystyle x-3=2Ax+2A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 2A=1$
$\displaystyle A=\frac12$
$\displaystyle 2A+B=-3$
$\displaystyle 2\cdot\frac12+B=-3$
$\displaystyle B=-4$
$\displaystyle \text{Now, } \int \frac{x-3}{x^2+2x-4}\,dx$
$\displaystyle =\int\left[\frac{\frac12(2x+2)-4}{x^2+2x-4}\right]dx$
$\displaystyle =\frac12\int\frac{2x+2}{x^2+2x-4}\,dx-4\int\frac{dx}{x^2+2x-4}$
$\displaystyle =\frac12\int\frac{2x+2}{x^2+2x-4}\,dx-4\int\frac{dx}{(x+1)^2-5}$
$\displaystyle =\frac12\log|x^2+2x-4|-\frac{4}{2\sqrt5}\log\left|\frac{x+1-\sqrt5}{x+1+\sqrt5}\right|+C$
$\displaystyle =\frac12\log|x^2+2x-4|-\frac{2}{\sqrt5}\log\left|\frac{x+1-\sqrt5}{x+1+\sqrt5}\right|+C$
$\\$
$\displaystyle \textbf{Question 4: }~\int \frac{2x-3}{x^2+6x+13}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{2x-3}{x^2+6x+13}\,dx$
$\displaystyle 2x-3=A\frac{d}{dx}(x^2+6x+13)+B$
$\displaystyle 2x-3=A(2x+6)+B$
$\displaystyle 2x-3=2Ax+6A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 2A=2$
$\displaystyle A=1$
$\displaystyle 6A+B=-3$
$\displaystyle 6+B=-3$
$\displaystyle B=-9$
$\displaystyle \therefore\ 2x-3=1(2x+6)-9$
$\displaystyle \text{Now, } \int \frac{2x-3}{x^2+6x+13}\,dx$
$\displaystyle =\int \frac{2x+6-9}{x^2+6x+13}\,dx$
$\displaystyle =\int \frac{2x+6}{x^2+6x+13}\,dx-\int \frac{9\,dx}{x^2+6x+13}$
$\displaystyle =\int \frac{2x+6}{x^2+6x+13}\,dx-9\int \frac{dx}{x^2+6x+3^2-3^2+13}$
$\displaystyle =\int \frac{2x+6}{x^2+6x+13}\,dx-9\int \frac{dx}{(x+3)^2+2^2}$
$\displaystyle =\log|x^2+6x+13|-9\cdot\frac12\tan^{-1}\left(\frac{x+3}{2}\right)+C$
$\displaystyle =\log|x^2+6x+13|-\frac92\tan^{-1}\left(\frac{x+3}{2}\right)+C$
$\\$
$\displaystyle \textbf{Question 5: }~\int \frac{x-1}{3x^2-4x+3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x-1}{3x^2-4x+3}\,dx$
$\displaystyle x-1=A\frac{d}{dx}(3x^2-4x+3)+B$
$\displaystyle x-1=A(6x-4)+B$
$\displaystyle x-1=6Ax-4A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 6A=1$
$\displaystyle A=\frac16$
$\displaystyle B-4A=-1$
$\displaystyle B-\frac{4}{6}=-1$
$\displaystyle B=-1+\frac23$
$\displaystyle B=-\frac13$
$\displaystyle \text{Now, } \int \frac{x-1}{3x^2-4x+3}\,dx$
$\displaystyle =\int\left[\frac{\frac16(6x-4)-\frac13}{3x^2-4x+3}\right]dx$
$\displaystyle =\frac16\int\frac{6x-4}{3x^2-4x+3}\,dx-\frac13\int\frac{dx}{3x^2-4x+3}$
$\displaystyle =\frac16\int\frac{6x-4}{3x^2-4x+3}\,dx-\frac13\int\frac{dx}{3\left(x^2-\frac43x+1\right)}$
$\displaystyle =\frac16\int\frac{6x-4}{3x^2-4x+3}\,dx-\frac19\int\frac{dx}{\left(x-\frac23\right)^2+\left(\frac{\sqrt5}{3}\right)^2}$
$\displaystyle =\frac16\log|3x^2-4x+3|-\frac19\cdot\frac{3}{\sqrt5}\tan^{-1}\left(\frac{x-\frac23}{\frac{\sqrt5}{3}}\right)+C$
$\displaystyle =\frac16\log|3x^2-4x+3|-\frac{1}{3\sqrt5}\tan^{-1}\left(\frac{3x-2}{\sqrt5}\right)+C$
$\displaystyle =\frac16\log|3x^2-4x+3|-\frac{\sqrt5}{15}\tan^{-1}\left(\frac{3x-2}{\sqrt5}\right)+C$
$\\$
$\displaystyle \textbf{Question 6: }~\int \frac{2x}{2+x-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{2x}{2+x-x^2}\,dx$
$\displaystyle 2x=A\frac{d}{dx}(2+x-x^2)+B$
$\displaystyle 2x=A(0+1-2x)+B$
$\displaystyle 2x=(-2A)x+A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle -2A=2$
$\displaystyle A=-1$
$\displaystyle A+B=0$
$\displaystyle -1+B=0$
$\displaystyle B=1$
$\displaystyle \text{Now, } \int \frac{2x}{2+x-x^2}\,dx$
$\displaystyle =\int \frac{-1(1-2x)+1}{-x^2+x+2}\,dx$
$\displaystyle =-\int \frac{1-2x}{-x^2+x+2}\,dx+\int \frac{dx}{-x^2+x+2}$
$\displaystyle =-I_1+I_2\ \text{(say) where}$
$\displaystyle I_1=\int \frac{1-2x}{-x^2+x+2}\,dx$
$\displaystyle I_2=\int \frac{dx}{-x^2+x+2}$
$\displaystyle I_1=\int \frac{1-2x}{-x^2+x+2}\,dx$
$\displaystyle \text{Let }-x^2+x+2=t$
$\displaystyle \Rightarrow (1-2x)\,dx=dt$
$\displaystyle I_1=\int \frac{dt}{t}$
$\displaystyle I_1=\log|t|+C_1$
$\displaystyle I_1=\log|2+x-x^2|+C_1\quad(2)$
$\displaystyle I_2=\int \frac{dx}{-x^2+x+2}$
$\displaystyle I_2=-\int \frac{dx}{x^2-x-2}$
$\displaystyle I_2=-\int \frac{dx}{x^2-x+\left(\frac12\right)^2-\left(\frac12\right)^2-2}$
$\displaystyle I_2=-\int \frac{dx}{\left(x-\frac12\right)^2-\left(\frac32\right)^2}$
$\displaystyle I_2=-\frac{1}{2\cdot\frac32}\log\left|\frac{x-\frac12-\frac32}{x-\frac12+\frac32}\right|+C_2$
$\displaystyle I_2=-\frac13\log\left|\frac{x-2}{x+1}\right|+C_2\quad(3)$
$\displaystyle \text{From (1), (2) and (3)}$
$\displaystyle \int \frac{2x}{2+x-x^2}\,dx=-\log|2+x-x^2|-\frac13\log\left|\frac{x-2}{x+1}\right|+C_1+C_2$
$\displaystyle =-\log|2+x-x^2|+\frac13\log\left|\frac{x+1}{x-2}\right|+C$
$\displaystyle \text{where }C=C_1+C_2$
$\\$
$\displaystyle \textbf{Question 7: }~\int \frac{1-3x}{3x^2+4x+2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{1-3x}{3x^2+4x+2}\,dx$
$\displaystyle 1-3x=A\frac{d}{dx}(3x^2+4x+2)+B$
$\displaystyle 1-3x=A(6x+4)+B$
$\displaystyle 1-3x=6Ax+4A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 6A=-3$
$\displaystyle A=-\frac12$
$\displaystyle 4A+B=1$
$\displaystyle 4\left(-\frac12\right)+B=1$
$\displaystyle B=3$
$\displaystyle 1-3x=-\frac12(6x+4)+3$
$\displaystyle \text{Now, } \int \frac{1-3x}{3x^2+4x+2}\,dx$
$\displaystyle =\int\left[\frac{-\frac12(6x+4)+3}{3x^2+4x+2}\right]dx$
$\displaystyle =-\frac12\int\frac{6x+4}{3x^2+4x+2}\,dx+3\int\frac{dx}{3x^2+4x+2}$
$\displaystyle =-\frac12 I_1+3I_2\ \text{(say)}$
$\displaystyle I_1=\int\frac{6x+4}{3x^2+4x+2}\,dx$
$\displaystyle \text{Let }3x^2+4x+2=t$
$\displaystyle \Rightarrow (6x+4)\,dx=dt$
$\displaystyle I_1=\int\frac{dt}{t}$
$\displaystyle I_1=\log|t|+C_1$
$\displaystyle I_1=\log|3x^2+4x+2|+C_1\quad(2)$
$\displaystyle I_2=\int\frac{dx}{3x^2+4x+2}$
$\displaystyle I_2=\frac13\int\frac{dx}{x^2+\frac43x+\frac23}$
$\displaystyle I_2=\frac13\int\frac{dx}{x^2+\frac43x+\left(\frac23\right)^2-\left(\frac23\right)^2+\frac23}$
$\displaystyle I_2=\frac13\int\frac{dx}{\left(x+\frac23\right)^2+\frac29}$
$\displaystyle I_2=\frac13\int\frac{dx}{\left(x+\frac23\right)^2+\left(\frac{\sqrt2}{3}\right)^2}$
$\displaystyle I_2=\frac13\cdot\frac{3}{\sqrt2}\tan^{-1}\left(\frac{x+\frac23}{\frac{\sqrt2}{3}}\right)+C_2$
$\displaystyle I_2=\frac1{\sqrt2}\tan^{-1}\left(\frac{3x+2}{\sqrt2}\right)+C_2\quad(3)$
$\displaystyle \text{From (1), (2) and (3)}$
$\displaystyle \int \frac{1-3x}{3x^2+4x+2}\,dx=-\frac12\log|3x^2+4x+2|+\frac3{\sqrt2}\tan^{-1}\left(\frac{3x+2}{\sqrt2}\right)+C$
$\\$
$\displaystyle \textbf{Question 8: }~\int \frac{2x+5}{x^2-x-2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{2x+5}{x^2-x-2}\,dx$
$\displaystyle 2x+5=A\frac{d}{dx}(x^2-x-2)+B$
$\displaystyle 2x+5=A(2x-1)+B$
$\displaystyle 2x+5=2Ax+B-A$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 2A=2$
$\displaystyle A=1$
$\displaystyle B-A=5$
$\displaystyle B-1=5$
$\displaystyle B=6$
$\displaystyle \therefore\ 2x+5=1(2x-1)+6$
$\displaystyle \text{Now, } \int \frac{2x+5}{x^2-x-2}\,dx$
$\displaystyle =\int\frac{(2x-1)+6}{x^2-x-2}\,dx$
$\displaystyle =\int\frac{2x-1}{x^2-x-2}\,dx+6\int\frac{dx}{x^2-x-2}$
$\displaystyle =I_1+6I_2\ \text{(say)}\quad(1)$
$\displaystyle I_1=\int\frac{2x-1}{x^2-x-2}\,dx$
$\displaystyle \text{Let }x^2-x-2=t$
$\displaystyle \Rightarrow (2x-1)\,dx=dt$
$\displaystyle I_1=\int\frac{dt}{t}$
$\displaystyle I_1=\log|t|+C_1$
$\displaystyle I_1=\log|x^2-x-2|+C_1\quad(2)$
$\displaystyle I_2=\int\frac{dx}{x^2-x-2}$
$\displaystyle I_2=\int\frac{dx}{x^2-x+\left(\frac12\right)^2-\left(\frac12\right)^2-2}$
$\displaystyle I_2=\int\frac{dx}{\left(x-\frac12\right)^2-\left(\frac32\right)^2}$
$\displaystyle I_2=\frac1{2\cdot\frac32}\log\left|\frac{x-\frac12-\frac32}{x-\frac12+\frac32}\right|+C_2$
$\displaystyle I_2=\frac13\log\left|\frac{x-2}{x+1}\right|+C_2\quad(3)$
$\displaystyle \text{From (1), (2) and (3)}$
$\displaystyle \int \frac{2x+5}{x^2-x-2}\,dx=\log|x^2-x-2|+\frac63\log\left|\frac{x-2}{x+1}\right|+C_1+C_2$
$\displaystyle =\log|x^2-x-2|+2\log\left|\frac{x-2}{x+1}\right|+C$
$\\$
$\displaystyle \textbf{Question 9: }~\int \frac{ax^3+bx}{x^4+c^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{ax^3+bx}{x^4+c^2}\,dx$
$\displaystyle =\int \frac{ax^3}{x^4+c^2}\,dx+\int \frac{bx}{(x^2)^2+c^2}\,dx$
$\displaystyle =I_1+I_2\ \text{(say)}$
$\displaystyle \text{Where } I_1=\int \frac{ax^3}{x^4+c^2}\,dx \text{ and } I_2=\int \frac{bx}{(x^2)^2+c^2}\,dx$
$\displaystyle \text{Now, } I_1=\int \frac{ax^3}{x^4+c^2}\,dx$
$\displaystyle \text{Let } x^4+c^2=t$
$\displaystyle \Rightarrow 4x^3\,dx=dt$
$\displaystyle \Rightarrow x^3\,dx=\frac{dt}{4}$
$\displaystyle I_1=\frac{a}{4}\int \frac{dt}{t}$
$\displaystyle I_1=\frac{a}{4}\log|t|+C_1$
$\displaystyle I_1=\frac{a}{4}\log|x^4+c^2|+C_1$
$\displaystyle \text{Now, } I_2=\int \frac{bx}{(x^2)^2+c^2}\,dx$
$\displaystyle \text{Let } x^2=p$
$\displaystyle \Rightarrow 2x\,dx=dp$
$\displaystyle \Rightarrow x\,dx=\frac{dp}{2}$
$\displaystyle I_2=\frac{b}{2}\int \frac{dp}{p^2+c^2}$
$\displaystyle I_2=\frac{b}{2}\cdot\frac{1}{c}\tan^{-1}\left(\frac{p}{c}\right)+C_2$
$\displaystyle I_2=\frac{b}{2c}\tan^{-1}\left(\frac{x^2}{c}\right)+C_2$
$\displaystyle \therefore \int \frac{ax^3+bx}{x^4+c^2}\,dx=\frac{a}{4}\log|x^4+c^2|+\frac{b}{2c}\tan^{-1}\left(\frac{x^2}{c}\right)+C$
$\\$ s
$\displaystyle \textbf{Question 10: }~\int \frac{(3\sin x-2)\cos x}{5-\cos^2 x-4\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{(3\sin x-2)\cos x}{5\cos^2 x-4\sin x}\,dx$
$\displaystyle =\int \frac{(3\sin x-2)\cos x}{5(1-\sin^2 x)-4\sin x}\,dx$
$\displaystyle =\int \frac{(3\sin x-2)\cos x}{\sin^2 x-4\sin x+4}\,dx$
$\displaystyle \text{Let } \sin x=t$
$\displaystyle \Rightarrow \cos x\,dx=dt$
$\displaystyle \int \frac{(3t-2)\,dt}{t^2-4t+4}$
$\displaystyle 3t-2=A\frac{d}{dt}(t^2-4t+4)+B$
$\displaystyle 3t-2=A(2t-4)+B$
$\displaystyle 3t-2=2At-4A+B$
$\displaystyle \text{Comparing coefficients of like powers of }t$
$\displaystyle 2A=3$
$\displaystyle A=\frac32$
$\displaystyle B-4A=-2$
$\displaystyle B-4\cdot\frac32=-2$
$\displaystyle B=4$
$\displaystyle \therefore\ 3t-2=\frac32(2t-4)+4$
$\displaystyle \int \frac{(3t-2)\,dt}{t^2-4t+4}=\int \frac{\frac32(2t-4)+4}{t^2-4t+4}\,dt$
$\displaystyle =\frac32\int\frac{2t-4}{t^2-4t+4}\,dt+4\int\frac{dt}{t^2-4t+4}$
$\displaystyle =\frac32 I_1+4 I_2\ \text{(say)}\quad(1)$
$\displaystyle I_1=\int\frac{2t-4}{t^2-4t+4}\,dt$
$\displaystyle \text{Let } t^2-4t+4=p$
$\displaystyle \Rightarrow (2t-4)\,dt=dp$
$\displaystyle I_1=\int\frac{dp}{p}$
$\displaystyle I_1=\log|p|+C_1$
$\displaystyle I_1=\log|t^2-4t+4|+C_1\quad(2)$
$\displaystyle I_2=\int\frac{dt}{t^2-4t+4}$
$\displaystyle I_2=\int\frac{dt}{(t-2)^2}$
$\displaystyle I_2=\int (t-2)^{-2}\,dt$
$\displaystyle I_2=\frac{(t-2)^{-1}}{-1}+C_2$
$\displaystyle I_2=-\frac{1}{t-2}+C_2\quad(3)$
$\displaystyle \text{From (1), (2) and (3)}$
$\displaystyle \int \frac{(3\sin x-2)\cos x}{5\cos^2 x-4\sin x}\,dx=\frac32\log|t^2-4t+4|-\frac{4}{t-2}+C$
$\displaystyle =\frac32\log|\sin^2 x-4\sin x+4|-\frac{4}{2-\sin x}+C$
$\displaystyle =\frac32\log|(\sin x-2)^2|-\frac{4}{2-\sin x}+C$
$\displaystyle =3\log|2-\sin x|+\frac{4}{2-\sin x}+C$
$\\$
$\displaystyle \textbf{Question 11: }~\int \frac{x+2}{2x^2+6x+5}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x+2}{2x^2+6x+5}\,dx$
$\displaystyle x+2=A\frac{d}{dx}(2x^2+6x+5)+B$
$\displaystyle x+2=A(4x+6)+B$
$\displaystyle x+2=4Ax+6A+B$
$\displaystyle \text{Comparing coefficients of like powers of }x$
$\displaystyle 4A=1$
$\displaystyle A=\frac14$
$\displaystyle 6A+B=2$
$\displaystyle 6\cdot\frac14+B=2$
$\displaystyle B=\frac12$
$\displaystyle \text{Now, } \int \frac{x+2}{2x^2+6x+5}\,dx$
$\displaystyle =\int\left[\frac{\frac14(4x+6)+\frac12}{2x^2+6x+5}\right]dx$
$\displaystyle =\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac12\int\frac{dx}{2x^2+6x+5}$
$\displaystyle =\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac14\int\frac{dx}{x^2+3x+\frac52}$
$\displaystyle =\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac14\int\frac{dx}{x^2+3x+\left(\frac32\right)^2-\left(\frac32\right)^2+\frac52}$
$\displaystyle =\frac14\int\frac{4x+6}{2x^2+6x+5}\,dx+\frac14\int\frac{dx}{\left(x+\frac32\right)^2+\frac14}$
$\displaystyle =\frac14\log|2x^2+6x+5|+\frac14\cdot2\tan^{-1}\left(\frac{x+\frac32}{\frac12}\right)+C$
$\displaystyle =\frac14\log|2x^2+6x+5|+\frac12\tan^{-1}(2x+3)+C$
$\\$
$\displaystyle \textbf{Question 12: }~\int \frac{5x-2}{1+2x+3x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{5x-2}{1+2x+3x^2}\,dx$
$\displaystyle =\int \frac{5x-2}{3x^2+2x+1}\,dx$
$\displaystyle \text{We express }5x-2=A\frac{d}{dx}(3x^2+2x+1)+B$
$\displaystyle 5x-2=A(6x+2)+B$
$\displaystyle \text{Equating coefficients of }x\text{ and constants, we get}$
$\displaystyle 6A=5 \text{ and } 2A+B=-2$
$\displaystyle A=\frac56 \text{ and } B=-\frac{11}{3}$
$\displaystyle \therefore I=\int \frac{\frac56(6x+2)-\frac{11}{3}}{3x^2+2x+1}\,dx$
$\displaystyle =\frac56\int \frac{6x+2}{3x^2+2x+1}\,dx-\frac{11}{3}\int \frac{1}{3x^2+2x+1}\,dx$
$\displaystyle =\frac56 I_1-\frac{11}{3}I_2\ \text{(1)}$
$\displaystyle \text{Now } I_1=\int \frac{6x+2}{3x^2+2x+1}\,dx$
$\displaystyle \text{Let }3x^2+2x+1=t$
$\displaystyle \text{On differentiating both sides, we get}$
$\displaystyle (6x+2)\,dx=dt$
$\displaystyle \therefore I_1=\int \frac{1}{t}\,dt$
$\displaystyle I_1=\log|t|+C_1$
$\displaystyle I_1=\log|3x^2+2x+1|+C_1\ \text{(2)}$
$\displaystyle \text{And, } I_2=\int \frac{1}{3x^2+2x+1}\,dx$
$\displaystyle =\frac13\int \frac{1}{x^2+\frac23x+\frac13}\,dx$
$\displaystyle =\frac13\int \frac{1}{x^2+\frac23x+\frac19-\frac19+\frac13}\,dx$
$\displaystyle =\frac13\int \frac{1}{\left(x+\frac13\right)^2+\frac29}\,dx$
$\displaystyle \text{Let }x+\frac13=t$
$\displaystyle \text{On differentiating both sides, we get }dx=dt$
$\displaystyle \therefore I_2=\frac13\int \frac{1}{t^2+\left(\frac{\sqrt2}{3}\right)^2}\,dt$
$\displaystyle =\frac13\cdot\frac{3}{\sqrt2}\tan^{-1}\left(\frac{3t}{\sqrt2}\right)+C_2$
$\displaystyle =\frac1{\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C_2\ \text{(3)}$
$\displaystyle \text{From (1), (2) and (3), we get}$
$\displaystyle I=\frac56\log|3x^2+2x+1|-\frac{11}{3}\left(\frac1{\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)\right)+C$
$\displaystyle \text{Hence, } \int \frac{5x-2}{1+2x+3x^2}\,dx=\frac56\log|3x^2+2x+1|-\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C$
$\\$
$\displaystyle \textbf{Question 13: }~\int \frac{x+5}{3x^2+13x-10}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{x+5}{3x^2+13x-10}\,dx$
$\displaystyle =\int \frac{x+5}{3x^2+15x-2x-10}\,dx$
$\displaystyle =\int \frac{x+5}{3x(x+5)-2(x+5)}\,dx$
$\displaystyle =\int \frac{x+5}{(3x-2)(x+5)}\,dx$
$\displaystyle =\int \frac{1}{3x-2}\,dx$
$\displaystyle I=\frac13\log|3x-2|+C$
$\\$
$\displaystyle \textbf{Question 14: }~\int \frac{(3\sin x-2)\cos x}{13-\cos^2 x-7\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{(3\sin x-2)\cos x}{13-\cos^2 x-7\sin x}\,dx$
$\displaystyle =\int \frac{(3\sin x-2)\cos x}{13-(1-\sin^2 x)-7\sin x}\,dx$
$\displaystyle =\int \frac{(3\sin x-2)\cos x}{\sin^2 x-7\sin x+12}\,dx$
$\displaystyle =\int \frac{(3\sin x-2)\cos x}{(\sin x-3)(\sin x-4)}\,dx$
$\displaystyle \text{Let } \sin x=t$
$\displaystyle \Rightarrow \cos x\,dx=dt$
$\displaystyle \therefore I=\int \frac{3t-2}{(t-3)(t-4)}\,dt$
$\displaystyle \text{Using partial fractions, we get}$
$\displaystyle \frac{3t-2}{(t-3)(t-4)}=\frac{A}{t-3}+\frac{B}{t-4}$
$\displaystyle 3t-2=A(t-4)+B(t-3)$
$\displaystyle 3t-2=(A+B)t-4A-3B$
$\displaystyle \text{Comparing coefficients, we get}$
$\displaystyle A+B=3$
$\displaystyle -4A-3B=-2$
$\displaystyle A=-7,\ B=10$
$\displaystyle \therefore I=-7\int \frac{dt}{t-3}+10\int \frac{dt}{t-4}$
$\displaystyle I=-7\log|t-3|+10\log|t-4|+C$
$\displaystyle I=-7\log|\sin x-3|+10\log|\sin x-4|+C$
$\\$
$\displaystyle \textbf{Question 15: }~\int \frac{x+7}{3x^2+25x+28}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{x+7}{3x^2+25x+28}\,dx$
$\displaystyle =\int \frac{x+7}{3x^2+21x+4x+28}\,dx$
$\displaystyle =\int \frac{x+7}{3x(x+7)+4(x+7)}\,dx$
$\displaystyle =\int \frac{x+7}{(3x+4)(x+7)}\,dx$
$\displaystyle =\int \frac{1}{3x+4}\,dx$
$\displaystyle I=\frac13\log|3x+4|+C$
$\\$
$\displaystyle \textbf{Question 16: }~\int \frac{x^3}{x^4+x^2+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{x^3}{x^4+x^2+1}\,dx$
$\displaystyle =\int \frac{x^2\cdot x}{(x^2)^2+x^2+1}\,dx$
$\displaystyle \text{Let } x^2=t \text{ or } 2x\,dx=dt$
$\displaystyle \Rightarrow I=\frac12\int \frac{t}{t^2+t+1}\,dt$
$\displaystyle =\frac14\int \frac{2t}{t^2+t+1}\,dt$
$\displaystyle =\frac14\int \frac{(2t+1)-1}{t^2+t+1}\,dt$
$\displaystyle =\frac14\left[\int \frac{2t+1}{t^2+t+1}\,dt-\int \frac{1}{t^2+t+1}\,dt\right]$
$\displaystyle =\frac14\left[\log|t^2+t+1|-\int \frac{1}{t^2+t+\frac14+\frac34}\,dt\right]$
$\displaystyle =\frac14\left[\log|t^2+t+1|-\int \frac{1}{\left(t+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\,dt\right]$
$\displaystyle =\frac14\left[\log|t^2+t+1|-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{t+\frac12}{\frac{\sqrt3}{2}}\right)\right]+C$
$\displaystyle =\frac14\left[\log|t^2+t+1|-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2t+1}{\sqrt3}\right)\right]+C$
$\displaystyle =\frac14\left[\log|x^4+x^2+1|-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x^2+1}{\sqrt3}\right)\right]+C$
$\\$
$\displaystyle \textbf{Question 17: }~\int \frac{x^3-3x}{x^4+2x^2-4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{x^3-3x}{x^4+2x^2-4}\,dx$
$\displaystyle =\int \frac{x(x^2-3)}{x^4+2x^2-4}\,dx$
$\displaystyle \text{Let } x^2=t,\ \text{or }2x\,dx=dt$
$\displaystyle \Rightarrow I=\frac12\int \frac{t-3}{t^2+2t-4}\,dt$
$\displaystyle =\frac14\int \frac{2t-6}{t^2+2t-4}\,dt$
$\displaystyle =\frac14\int \frac{(2t+2)-8}{t^2+2t-4}\,dt$
$\displaystyle =\frac14\left[\int \frac{2t+2}{t^2+2t-4}\,dt-\int \frac{8}{t^2+2t-4}\,dt\right]$
$\displaystyle =\frac14(I_1+I_2)\quad\text{(i)}$
$\displaystyle \text{Now, } I_1=\int \frac{2t+2}{t^2+2t-4}\,dt$
$\displaystyle \text{Let } t^2+2t-4=u$
$\displaystyle \Rightarrow (2t+2)\,dt=du$
$\displaystyle \Rightarrow I_1=\int \frac{du}{u}=\log|u|+C_1$
$\displaystyle I_1=\log|t^2+2t-4|+C_1$
$\displaystyle \text{Now, } I_2=\int \frac{-8}{t^2+2t-4}\,dt$
$\displaystyle =-8\int \frac{dt}{(t+1)^2-5}$
$\displaystyle I_2=-\frac{8}{2\sqrt5}\log\left|\frac{t+1-\sqrt5}{t+1+\sqrt5}\right|+C_2$
$\displaystyle I_2=-\frac{4}{\sqrt5}\log\left|\frac{t+1-\sqrt5}{t+1+\sqrt5}\right|+C_2$
$\displaystyle \text{So, from (i), we get}$
$\displaystyle I=\frac14\log|t^2+2t-4|-\frac{4}{\sqrt5}\log\left|\frac{t+1-\sqrt5}{t+1+\sqrt5}\right|+C$
$\displaystyle I=\frac14\log|x^4+2x^2-4|-\frac{4}{\sqrt5}\log\left|\frac{x^2+1-\sqrt5}{x^2+1+\sqrt5}\right|+C$
$\\$