Question

Evaluate the following integrals:

$\int \frac{x}{\left(x^2+1\right)\left(x-1\right)}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x}{\left(x^2+1\right)\left(x-1\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x}{\left(x^2+1\right)\left(x-1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1} \\ \Rightarrow x=A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2,x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 0=A+B\mathrm{and}1=-B+C\mathrm{and}0=A-C \\ \mathrm{or}A=\frac{1}{2}\mathrm{and}B=-\frac{1}{2}\mathrm{and}C=\frac{1}{2} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{1}{2}}}{x-1}+\frac{-{\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}}}{x^2+1}\right)\mathrm{d}x \\ =\frac{1}{2}\int \frac{1}{x-1}\mathrm{d}x-\frac{1}{2}\int \frac{x}{x^2+1}\mathrm{d}x+\frac{1}{2}\int \frac{1}{x^2+1}\mathrm{d}x \\ =\frac{1}{2}{I}_1-\frac{1}{2}{I}_2+\frac{1}{2}{I}_3...\left(1\right) \\ \mathrm{Now},I_1=\int \frac{1}{x-1}\mathrm{d}x \\ \mathrm{Let}x-1=u \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x=\mathrm{d}u \\ \therefore I_1=\int \frac{1}{u}\mathrm{d}u \\ =\log \left|u\right|+c_1 \\ =\log \left|x-1\right|+c_1...\left(2\right) \\ \mathrm{And},I_2=\int \frac{x}{x^2+1}\mathrm{d}x \\ \mathrm{Let}\left(x^2+1\right)=u \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 2x\mathrm{d}x=\mathrm{d}u \\ \therefore I_2=\frac{1}{2}\int \frac{1}{u}\mathrm{d}u \\ =\frac{1}{2}\log \left|u\right|+c_2 \\ =\frac{1}{2}\log \left|x^2+1\right|+c_2...\left(3\right) \\ \mathrm{And},I_3=\int \frac{1}{x^2+1}\mathrm{d}x \\ ={\tan }^{-1}x+c_3...\left(4\right) \\ \mathrm{From}\left(1\right),\left(2\right),\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get} \\ \therefore I=\frac{1}{2}\left(\log \left|x-1\right|+c_1\right)-\frac{1}{2}\left(\frac{1}{2}\log \left|x^2+1\right|+c_2\right)+\frac{1}{2}\left({\tan }^{-1}x+c_3\right) \\ =\frac{1}{2}\log \left|x-1\right|-\frac{1}{4}\log \left|x^2+1\right|+\frac{1}{2}{\tan }^{-1}x+c \\ \mathrm{Hence},\int \frac{x}{\left(x^2+1\right)\left(x-1\right)}\mathrm{d}x=\frac{1}{2}\log \left|x-1\right|-\frac{1}{4}\log \left|x^2+1\right|+\frac{1}{2}{\tan }^{-1}x+c \end{gathered}$$