$\int \frac{x}{x^3-3 x+2} d x=\int \frac{x}{(x-1)^2} \frac{d x}{(x+2)}$
Now, by partial fraction method
$$
\begin{aligned}
& \frac{x}{(x-1)^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)} \\
& \Rightarrow x=A(x-1)(x+2)+B(x+2)+C(x-1)^2
\end{aligned}
$$
On comparing the coefficient of different terms, we are getting
$$
A=\frac{2}{9}, B=\frac{1}{3} \text { and } C=-\frac{2}{9}
$$
So, $\int \frac{x}{x^3-3 x+2} d x=\frac{2}{9} \int \frac{d x}{x-1}+\frac{1}{3} \int \frac{d x}{(x-1)^2}$
$$
-\frac{2}{9} \int \frac{d x}{x+2}
$$
$=\frac{2}{9} \log |x-1|-\frac{1}{3} \cdot \frac{1}{(x-1)}-\frac{2}{9} \log |x+2|+c$
$=-\frac{1}{3} \cdot \frac{1}{(x-1)}+\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+c$