To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$

Question

To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$

$\frac{1}{x^2(x+a)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{(x+a)^2}$

One method of finding the values of the constants A,B,C & D is as follows.

$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1$

$Ax(x^2+a^2+2ax)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$

$A(x^3+a^2x+2ax^2)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$

$x^3(A+C)+x^2(2Aa+B+Ca+D)+x(Aa^2+2Ba)+Ba^2=1$

Equating coeffecients of $x^3,x^2,x^1,x^0$

$A+C=0$; $2Aa+B+Ca+D=0$; $Aa^2+2Ba=0$; $Ba^2=1$

$B=\frac{1}{a^2}$

$A=\frac{-2}{a^3}$

$C=\frac{2}{a^3}$

$D=\frac{1}{a^2}$

Is there a shorter method to find the values of A,B,C & D ?

Answer 1

$$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1$$

$x=0$ yields $b= \frac{1}{a^2}$.

$x=-a$ yields $d= \frac{1}{a^2}$.

Then you can derivate and set again $x=-a$ and $x=0$ to get $A,C$ or replace $B,D$ in that equation to get

$$Ax(x+a)^2+C(x^2)(x+a)+\frac{2x^2}{a^2}+\frac{2ax}{a^2}+1=1$$

After canceling the 1's, you can divide by $x(x+a)$ and get

$$A(x+a)+Cx+\frac{2}{a^2}=0$$

$x=0$ yields $A=-\frac{2}{a^3}$

and $x=-a$ yields $C=\frac{2}{a^3}$

Answer 2

In this equation: $$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1 $$ one little thing you can do is to evaluate at $x = -a$ to get $0 +0 +0 +D(-a)^2 = 1$ to get $D$. After that can evaluate at $x=0$ to get $B$.