$\int_{0}^{2}\frac{1}{(x+2)(x-1)}$
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.