Let $\frac{x}{\left(x^2+1\right)(x-1)}=\frac{A}{x-1}+\frac{B x+C}{x^2+1}$ $\Rightarrow \quad x=\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x-1)$
Put $x=1,0$
$\begin{aligned}
&\Rightarrow \mathrm{A}=\frac{1}{2} ; \mathrm{C}=\frac{1}{2} \Rightarrow \mathrm{B}=-\frac{1}{2} \\
&\therefore \mathrm{I}=\frac{1}{2} \int \frac{d x}{x-1}-\frac{1}{2} \int \frac{x}{x^2+1} d x+\frac{1}{2} \int \frac{d x}{x^2+1} \\
&=\frac{1}{2} \log (x-1)-\frac{1}{4} \log \left(x^2+1\right)+\frac{1}{2} \tan ^{-1} x+\mathrm{C}
\end{aligned}$