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 a good start!
I'd suggest trying to reduce the number of variables. You can solve the first equation for $B=1-A$. Then try multiplying the second equation by 3 to get $12A-3B+3C=0$. Add this to the 3rd equation. Combine with the substitution for $B$ and solve for $A$. Then you can work backwards through the equations to solve for the other variables.
For a more systematic way to solve such things you may be interested in studying linear algebra.
