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

Question

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

Answer 1

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.

Answer 2

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 $A+B=1$, $A=\frac9{25}$.

Answer 3

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 $x^2$:

$$1=A+B\implies B=\frac{16}{25}$$

Answer 4

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}

Answer 5

$$B=1-A$$

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.

$$4A-6+6A-3+15A=0\\\implies 25A=9\\\implies A=\frac9{25}$$