Evaluating $\int_{0}^{2}\frac{1}{(x+2)(x-1)}dx$ $\int_{0}^{2}\frac{1}{(x+2)(x-1)}$
Will I be able to use partial fractions on the indefinite integral and then evaluate my answer as f(2)-f(0)?
Any help is appreciated as always.  :)

EDIT:  I looked this up and apparently the integral doesn't converge so now I need help finding where I went wrong.
I had solved for  1 = $\frac{A}{x+2} + \frac{B}{x-1}$
sub in x = 1   and $B=\frac{1}{3}$
sub in x=-2 and $A=\frac{-1}{3}$
Giving:
$\frac{-1}{3}\int\frac{dx}{x+2} + \frac{1}{3}\int\frac{dx}{x-1}$
Simple u-substitution gives:
$\frac{-1}{3}ln|x+2|$  + $\frac{1}{3}ln|x-1|  +   C   $
Which I was going to then evaluate f(2)-f(0)  but somewhere I went wrong.

Hint 1:
$$\frac{1}{(x+2)(x-1)}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$$
Hint 2: Does this converge or diverge?  Think about $$\int_{-1}^{1}\frac{1}{x}dx.$$  Does the value of this integral even make sense?
