Let $\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}$ $\Rightarrow \quad x \equiv \mathrm{A}(x-1)(x+2)+\mathrm{B}(x+2)+\mathrm{C}(x-1)^2 \quad \ldots(i)$
Put $x=1, x=-2$ in (i), we get: $\mathrm{B}=\frac{1}{3} \& \mathrm{C}=\frac{-2}{9}$
$\begin{aligned}
&\therefore \mathrm{I}=\frac{2}{9} \int \frac{1}{x-1} d x+\frac{1}{3} \int \frac{1}{(x-1)^2} d x-\frac{2}{9} \int \frac{1}{x+2} d x \\
&=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+\mathrm{C}
\end{aligned}$