\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{x^2+x+1}{x^2-x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \left(\frac{x^2+x+1}{x^2-x}\right)\,dx
\displaystyle \frac{x^2+x+1}{x^2-x}=1+\frac{2x+1}{x^2-x}
\displaystyle \therefore \int \left(\frac{x^2+x+1}{x^2-x}\right)\,dx
\displaystyle =\int \left(1+\frac{2x+1}{x^2-x}\right)\,dx
\displaystyle =\int 1+\frac{2x-1+2}{x^2-x}\,dx
\displaystyle =\int dx+\int \frac{2x-1}{x^2-x}\,dx+\int \frac{2\,dx}{x^2-x+\left(\frac12\right)^2-\left(\frac12\right)^2}
\displaystyle =\int dx+\int \frac{2x-1}{x^2-x}\,dx+2\int \frac{dx}{\left(x-\frac12\right)^2-\left(\frac12\right)^2}
\displaystyle =x+\log|x^2-x|+2\cdot\frac{1}{2\cdot\frac12}\log\left|\frac{x-\frac12-\frac12}{x-\frac12+\frac12}\right|+C
\displaystyle =x+\log|x^2-x|+2\log\left|\frac{x-1}{x}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{x^2+x-1}{x^2+x-6}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \left(\frac{x^2+x-1}{x^2+x-6}\right)\,dx
\displaystyle \frac{x^2+x-1}{x^2+x-6}=1+\frac{5}{x^2+x-6}
\displaystyle \therefore \int \left(\frac{x^2+x-1}{x^2+x-6}\right)\,dx
\displaystyle =\int dx+5\int \frac{dx}{x^2+x-6}
\displaystyle =\int dx+5\int \frac{dx}{x^2+x+\left(\frac12\right)^2-\left(\frac12\right)^2-6}
\displaystyle =\int dx+5\int \frac{dx}{\left(x+\frac12\right)^2-\frac14-6}
\displaystyle =\int dx+5\int \frac{dx}{\left(x+\frac12\right)^2-\left(\frac52\right)^2}
\displaystyle =x+5\cdot\frac{1}{2\cdot\frac52}\log\left|\frac{x+\frac12-\frac52}{x+\frac12+\frac52}\right|+C
\displaystyle =x+\log\left|\frac{x-2}{x+3}\right|+C

\displaystyle \textbf{Question 3: }~\int \frac{1-x^2}{x(1-2x)}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle I=\int \frac{1-x^2}{x(1-2x)}\,dx
\displaystyle =\int \frac{-x^2+1}{-2x^2+x}\,dx
\displaystyle =\int \left(\frac12+\frac{1-\frac{x}{2}}{-2x^2+x}\right)\,dx
\displaystyle =\frac12\int dx+\frac12\int \frac{2-x}{-2x^2+x}\,dx
\displaystyle =\frac12\left[\int dx+\int \frac{2-x}{-2x^2+x}\,dx\right]
\displaystyle =\frac12[I_1+I_2]\ \text{(say)}
\displaystyle \text{where } I_1=\int dx \text{ and } I_2=\int \frac{2-x}{-2x^2+x}\,dx
\displaystyle \text{Now, } I_1=\int dx
\displaystyle =x+C_1
\displaystyle I_2=\int \frac{2-x}{-2x^2+x}\,dx
\displaystyle \text{Let } 2-x=A\frac{d}{dx}(-2x^2+x)+B
\displaystyle \Rightarrow 2-x=A(-4x+1)+B
\displaystyle \Rightarrow 2-x=-4Ax+A+B
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle -1=-4A
\displaystyle A=\frac14
\displaystyle A+B=2
\displaystyle \frac14+B=2
\displaystyle B=\frac74
\displaystyle \therefore I_2=\int \frac{\frac14(-4x+1)+\frac74}{-2x^2+x}\,dx
\displaystyle =\frac14\int \frac{-4x+1}{-2x^2+x}\,dx+\frac74\int \frac{dx}{-2x^2+x}
\displaystyle =\frac14\log|-2x^2+x|+\frac74\int \frac{dx}{-2x^2+x}+C_2
\displaystyle =\frac14\log|-2x^2+x|-\frac78\int \frac{dx}{x^2-\frac12x}
\displaystyle =\frac14\log|-2x^2+x|-\frac78\int \frac{dx}{\left(x-\frac14\right)^2-\left(\frac14\right)^2}
\displaystyle =\frac14\log|-2x^2+x|-\frac78\cdot\frac{1}{2\cdot\frac14}\log\left|\frac{x-\frac14-\frac14}{x-\frac14+\frac14}\right|+C_2
\displaystyle =\frac14\log|-2x^2+x|-\frac74\log\left|\frac{x-\frac12}{x}\right|+C_2
\displaystyle \text{Thus, } I=\frac12\left[x+\frac14\log|-2x^2+x|-\frac74\log\left|\frac{x-\frac12}{x}\right|\right]+C
\displaystyle =\frac12x+\log|x|-\frac34\log|1-2x|+C

\displaystyle \textbf{Question 4: }~\int \frac{x^2+1}{x^2-5x+6}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \left(\frac{x^2+1}{x^2-5x+6}\right)\,dx
\displaystyle \text{Dividing numerator by denominator}
\displaystyle \frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}\quad(1)
\displaystyle \text{Also } \frac{5x-5}{x^2-5x+6}=\frac{5x-5}{(x-2)(x-3)}
\displaystyle \text{Let } \frac{5x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}
\displaystyle \Rightarrow 5x-5=A(x-3)+B(x-2)
\displaystyle \text{Let } x=3
\displaystyle 5(3)-5=A(0)+B(1)
\displaystyle 10=B
\displaystyle \text{Let } x=2
\displaystyle 5(2)-5=A(-1)+B(0)
\displaystyle A=-5
\displaystyle \therefore \frac{5x-5}{(x-2)(x-3)}=-\frac{5}{x-2}+\frac{10}{x-3}\quad(2)
\displaystyle \text{From (1) and (2)}
\displaystyle I=\int dx-5\int \frac{dx}{x-2}+10\int \frac{dx}{x-3}
\displaystyle I=x-5\log|x-2|+10\log|x-3|+C

\displaystyle \textbf{Question 5: }~\int \frac{x^2}{x^2+7x+10}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^2}{x^2+7x+10}\,dx
\displaystyle \text{Now, divide numerator by denominator}
\displaystyle \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{x^2+7x+10}
\displaystyle \therefore \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{(x+2)(x+5)}\quad(1)
\displaystyle \text{Consider, } \frac{7x+10}{(x+2)(x+5)}=\frac{A}{x+2}+\frac{B}{x+5}
\displaystyle \Rightarrow 7x+10=A(x+5)+B(x+2)
\displaystyle \text{Let } x+5=0
\displaystyle \Rightarrow x=-5
\displaystyle 7(-5)+10=A(0)+B(-3)
\displaystyle -25=-3B
\displaystyle B=\frac{25}{3}
\displaystyle \text{Let } x+2=0
\displaystyle \Rightarrow x=-2
\displaystyle 7(-2)+10=A(3)+B(0)
\displaystyle -4=3A
\displaystyle A=-\frac{4}{3}
\displaystyle \therefore \frac{7x+10}{(x+2)(x+5)}=-\frac{4}{3(x+2)}+\frac{25}{3(x+5)}\quad(2)
\displaystyle \text{From (1) and (2)}
\displaystyle \frac{x^2}{x^2+7x+10}=1+\frac{4}{3(x+2)}-\frac{25}{3(x+5)}
\displaystyle \therefore \int \frac{x^2}{x^2+7x+10}\,dx=\int dx+\frac{4}{3}\int \frac{dx}{x+2}-\frac{25}{3}\int \frac{dx}{x+5}
\displaystyle =x+\frac{4}{3}\log|x+2|-\frac{25}{3}\log|x+5|+C

\displaystyle \textbf{Question 6: }~\int \frac{x^2+x+1}{x^2-x+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^2}{x^2+7x+10}\,dx
\displaystyle \text{Now,}
\displaystyle \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{x^2+7x+10}
\displaystyle \Rightarrow \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{x^2+2x+5x+10}
\displaystyle \Rightarrow \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{x(x+2)+5(x+2)}
\displaystyle \Rightarrow \frac{x^2}{x^2+7x+10}=1-\frac{7x+10}{(x+2)(x+5)}\quad(1)
\displaystyle \text{Consider, } \frac{7x+10}{(x+2)(x+5)}=\frac{A}{x+2}+\frac{B}{x+5}
\displaystyle \Rightarrow 7x+10=A(x+5)+B(x+2)
\displaystyle \text{Let } x+5=0
\displaystyle \Rightarrow x=-5
\displaystyle 7(-5)+10=A\cdot0+B(-5+2)
\displaystyle -25=-3B
\displaystyle B=\frac{25}{3}
\displaystyle \text{Let } x+2=0
\displaystyle \Rightarrow x=-2
\displaystyle 7(-2)+10=A(-2+5)+B\cdot0
\displaystyle -4=3A
\displaystyle A=-\frac{4}{3}
\displaystyle \therefore \frac{7x+10}{(x+2)(x+5)}=-\frac{4}{3(x+2)}+\frac{25}{3(x+5)}\quad(2)
\displaystyle \text{From (1) and (2)}
\displaystyle \frac{x^2}{x^2+7x+10}=1+\frac{4}{3(x+2)}-\frac{25}{3(x+5)}
\displaystyle \Rightarrow \int \frac{x^2}{x^2+7x+10}\,dx=\int dx+\frac{4}{3}\int \frac{dx}{x+2}-\frac{25}{3}\int \frac{dx}{x+5}
\displaystyle =x+\frac{4}{3}\log|x+2|-\frac{25}{3}\log|x+5|+C

\displaystyle \textbf{Question 7: }~\int \frac{(x-1)^2}{x^2+2x+2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \left(\frac{x^2+x+1}{x^2-x+1}\right)\,dx
\displaystyle \text{Now,}
\displaystyle \frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}
\displaystyle \therefore \int \left(\frac{x^2+x+1}{x^2-x+1}\right)\,dx=\int dx+\int \frac{2x}{x^2-x+1}\,dx
\displaystyle =\int dx+\int \frac{2x-1+1}{x^2-x+1}\,dx
\displaystyle =\int dx+\int \frac{2x-1}{x^2-x+1}\,dx+\int \frac{dx}{x^2-x+1}
\displaystyle =\int dx+\int \frac{2x-1}{x^2-x+1}\,dx+\int \frac{dx}{x^2-x+\left(\frac12\right)^2-\left(\frac12\right)^2+1}
\displaystyle =\int dx+\int \frac{2x-1}{x^2-x+1}\,dx+\int \frac{dx}{\left(x-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =x+\log|x^2-x+1|+\frac{2}{\sqrt3}\tan^{-1}\left(\frac{x-\frac12}{\frac{\sqrt3}{2}}\right)+C
\displaystyle =x+\log|x^2-x+1|+\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C

\displaystyle \textbf{Question 8: }~\int \frac{x^3+x^2+2x+1}{x^2-x+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{(x-1)^2}{x^2+2x+2}\,dx
\displaystyle =\int \frac{x^2-2x+1}{x^2+2x+2}\,dx
\displaystyle \text{Here,}
\displaystyle \frac{x^2-2x+1}{x^2+2x+2}=1-\frac{4x+1}{x^2+2x+2}\quad(1)
\displaystyle \text{Let } 4x+1=A\frac{d}{dx}(x^2+2x+2)+B
\displaystyle \Rightarrow 4x+1=A(2x+2)+B
\displaystyle \Rightarrow 4x+1=(2A)x+2A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 2A=4
\displaystyle A=2
\displaystyle 2A+B=1
\displaystyle 2\cdot2+B=1
\displaystyle B=-3
\displaystyle \therefore \int \frac{x^2-2x+1}{x^2+2x+2}\,dx
\displaystyle =\int dx-2\int \frac{2x+2}{x^2+2x+2}\,dx+3\int \frac{dx}{x^2+2x+2}
\displaystyle =x-2\log|x^2+2x+2|+3\int \frac{dx}{(x+1)^2+1^2}
\displaystyle =x-2\log|x^2+2x+2|+3\tan^{-1}(x+1)+C

\displaystyle \textbf{Question 9: }~\int \frac{x^2(x^4+4)}{x^2+4}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \left(\frac{x^3+x^2+2x+1}{x^2-x+1}\right)\,dx
\displaystyle \text{Dividing numerator by denominator}
\displaystyle \frac{x^3+x^2+2x+1}{x^2-x+1}=x+2+\frac{3x-1}{x^2-x+1}\quad(1)
\displaystyle \text{Let } 3x-1=A\frac{d}{dx}(x^2-x+1)+B
\displaystyle \Rightarrow 3x-1=A(2x-1)+B
\displaystyle \Rightarrow 3x-1=2Ax-A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 2A=3
\displaystyle A=\frac32
\displaystyle B-A=-1
\displaystyle B-\frac32=-1
\displaystyle B=\frac12
\displaystyle \therefore \int \left(\frac{x^3+x^2+2x+1}{x^2-x+1}\right)\,dx
\displaystyle =\int (x+2)\,dx+\int \frac{\frac32(2x-1)+\frac12}{x^2-x+1}\,dx
\displaystyle =\int (x+2)\,dx+\frac32\int \frac{2x-1}{x^2-x+1}\,dx+\frac12\int \frac{dx}{x^2-x+1}
\displaystyle =\int (x+2)\,dx+\frac32\int \frac{2x-1}{x^2-x+1}\,dx+\frac12\int \frac{dx}{x^2-x+\left(\frac12\right)^2-\left(\frac12\right)^2+1}
\displaystyle =\int (x+2)\,dx+\frac32\int \frac{2x-1}{x^2-x+1}\,dx+\frac12\int \frac{dx}{\left(x-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =\frac{x^2}{2}+2x+\frac32\log|x^2-x+1|+\frac{1}{\sqrt3}\tan^{-1}\left(\frac{x-\frac12}{\frac{\sqrt3}{2}}\right)+C
\displaystyle =\frac{x^2}{2}+2x+\frac32\log|x^2-x+1|+\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C

\displaystyle \textbf{Question 10: }~\int \frac{x^2}{x^2+6x+12}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^2(x^4+4)}{x^2+4}\,dx
\displaystyle =\int \frac{x^6+4x^2}{x^2+4}\,dx
\displaystyle \text{Now, dividing numerator by denominator}
\displaystyle \frac{x^6+4x^2}{x^2+4}=x^4-4x^2+20-\frac{80}{x^2+4}
\displaystyle \therefore I=\int \left(x^4-4x^2+20\right)\,dx-80\int \frac{dx}{x^2+4}
\displaystyle =\int x^4\,dx-4\int x^2\,dx+20\int dx-80\int \frac{dx}{x^2+2^2}
\displaystyle =\frac{x^5}{5}-4\cdot\frac{x^3}{3}+20x-80\cdot\frac12\tan^{-1}\left(\frac{x}{2}\right)+C
\displaystyle =\frac{x^5}{5}-\frac{4}{3}x^3+20x-40\tan^{-1}\left(\frac{x}{2}\right)+C

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