Question

Find :
$\int \frac{2}{(1-x)(1+x^2)}dx$

Answer 1

We have,

$I=\int \frac{2}{(1-x)(1+x^2)}dx$

$I=\int (\frac{1}{1-x}+\frac{x+1}{1+x^2})dx$

$I=-\int \left(\frac{1}{x-1}\right)dx+\frac{1}{2}\int \left(\frac{2x}{1+x^2}\right)dx+\int \left(\frac{1}{1+x^2}\right)dx$

$I=-\ln (x-1)+\frac{1}{2}\ln (x^2+1)+{\tan }^{-1}x+C$

Hence, this is the answer.