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

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

$2=A+A x^{2}+B x-B x^{2}+C-C x$

Equating the coefficient of $x^{2}, x$, and constant term, we obtain

$A-B=0$

$B-C=0$

$A+C=2$

On solving these equations, we obtain

$A=1, B=1$, and $C=1$

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

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

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

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