Evaluate the following integral:

$I=\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x$

$\frac{3}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}}$

$3=A\left(1+x^{2}\right)+(B x+C)(1-x)$

Equating similar terms

$A-B=0$

$B-C=0$

$A+C=3$

Solving

$A=\frac{3}{2}, B=\frac{3}{2}, C=\frac{3}{2}$

Thus,

$I=\frac{3}{2} \int \frac{d x}{1-x}+\frac{3}{2} \int \frac{x d x}{1+x^{2}}+\frac{3}{2} \int \frac{d x}{1+x^{2}}$

$=-\frac{3}{2} \log |1-x|+\frac{3}{2} \log \left|1+x^{2}\right|+\frac{3}{2} \tan ^{-1} x+C$

$I=\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x\right]+C$