Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions 
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$

I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C

You have\begin{align}-4&=0-4\\&=4A-B+C-4(A+B)\\&=-5B+C\end{align}and\begin{align}-4&=0-4\\&=4A-6B-3C-4(A+B)\\&=-10B-3C.\end{align}And now, you have\begin{align}4&=(-2)\times(-4)-4\\&=(-2)(-5B+C)-10B-3C\\&=-5C,\end{align}and therefore $C=-\frac45$. Now, since $-5B+C=-4$, you get that $B=\frac{16}{25}$. And, since $A+B=1$, $A=\frac9{25}$.
