As suggested in the title, I am given the following integral:

$$\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$$

I have tried to solve it several times, but I wind up with the wrong answer, although its close. I assume there is a blunder somewhere.

Applying long division, we have that 

$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{-3x-1}{x^2+x-2}$$

I note now that I may write this as 

$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{(-1)3x+1}{x^2+x-2} = 2x - 1 - \frac{3x+1}{x^2+x-2}$$

I did not notice this while doing the task originally, but I am having a hard time seeing how this will make answer different. (After calculating it on paper: it didn't.)

Thus, we have the integral 

$$\int 2x - 1 \space dx - \int \frac{3x+1}{x^2+x-2}\space dx$$

The first integral is trivial. For the second one, we have that 

$$\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$$

We have the set $A + B = 3, \space 2A - B = 1$.This gives us the solutions $A = \frac{4}{3}$ and $B = \frac{5}{3}$

Thus, we are left with 

$x^2 - x - \int \frac{3x+1}{x^2+x-2}\space dx = x^2 - x - \int \frac{4/3}{x-1} + \frac{5/3}{x+2} \space dx = x^2 - x - \frac{4}{3}\ln{|x-1|}-\frac{5}{3}\ln{|x+2|} + C$

I'm having a hard time finding my mistake here.

EDIT: Seems like I made my mistake in the long division part, confirming now.