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
 A: There is another way of finding $A$, $B$ and $C$ which is more efficient than forming and solving simultaneous equations.
Starting with the identity $$x^2=A(x+2)^2+B(x-3)(x+2)+C(x-3)$$
Choose values of $x$ to make the brackets zero.
So, putting $x=2$ gives $$4=C(-5)\implies C=-\frac45$$
Putting $x=3$ gives $$9=A(25)\implies A=\frac{9}{25}$$
Now that you’ve run out of convenient values of $x$ you can put $x=0$ or just look at coefficients, such as $x^2$:
$$1=A+B\implies B=\frac{16}{25}$$
A: 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.
A: 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 $A+B=1$, $A=\frac9{25}$.
A: From
$$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$
we evaluate  the coefficients directly as follows
\begin{align}
&A=\left[\frac{x^2}{(x+2)^2} - \frac{B(x-3)}{x+2} + \frac{C(x-3)}{(x+2)^2} \right]_{x=3}=\frac9{25}\\
&B=\left[\frac{x^2}{(x-3)(x+2)} - \frac{A(x+2)}{x-3} -{\frac C{x+2}}\right]_{x\to\infty}=1-A=\frac{16}{25}\\
 &C=\left[\frac{x^2}{x-3} - \frac{A(x+2)^2}{x-3} -{B(x+2)}\right]_{x=-2}=-\frac4{5}
\end{align}
A: $$B=1-A$$
Putting this in 2nd equation.
$$4A-1+A+C=0\\\implies C=1-5A$$
Now, we have both $B$ and $C$ in terms of $A$. Putting that in 3rd equation.
$$4A-6+6A-3+15A=0\\\implies 25A=9\\\implies A=\frac9{25}$$
