integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate    :

$\Large \int \frac{x}{x^3-3x+2}dx$

I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
Well I got A=1/3 and B=-1/3. as possible values. but that was not correct. I guess I missed something. 

You don't have all the components in the decomposition. It should be 
$$\frac {x}{(x-1)^2(x+2)} = \frac {Ax+C}{(x-1)^2} + \frac {B}{x+2}
=\frac {\frac29x+\frac19}{(x-1)^2} - \frac {\frac29}{x+2}$$
