Answer
$\dfrac{x^2}{(x-1)^2(x+1)^2}=\dfrac{1}{4(x-1)}+\dfrac{1}{4(x-1)^2}-\dfrac{1}{4(x+1)}+\dfrac{1}{4(x+1)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2}{(x-1)^2(x+1)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{x^2}{(x-1)^2(x+1)^2}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}+\dfrac{D}{(x+1)^2}$
Multiply all terms by the least common denominator $(x-1)^2(x+1)^2$:
$(x-1)^2(x+1)^2\cdot \dfrac{x^2}{(x-1)^2(x+1)^2}=(x-1)^2(x+1)^2\cdot \dfrac{A}{x-1}+(x-1)^2(x+1)^2\cdot \dfrac{B}{(x-1)^2}+(x-1)^2(x+1)^2\cdot \dfrac{C}{x+1}+(x-1)^2(x+1)^2\cdot \dfrac{D}{(x+1)^2}$
$x^2=A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2$
$x^2=A(x-1)(x^2+2x+1)-A+Bx+2Bx+B+C(x^2-2x+1)(x+1)+Dx^2-2Dx+D$
$x^2=A(x^3+2x^2+x-x^2-2x-1)+Bx^2+2Bx+B+C(x^3+x^2-2x^2-2x+x+1)+Dx^2-2Dx+D$
$x^2=Ax^3+Ax^2-Ax-A+Bx^2+2Bx+B+Cx^3-Cx^2-Cx+C+Dx^2-2Dx+D$
$x^2=(A+C)x^3+(A+B-C+D)x^2+(-A+2B-C-2D)x+(-A+B+C+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+C=0\\
A+B-C+D=1\\
-A+2B-C-2D=0\\
-A+B+C+D=0
\end{cases}$
Solve the system: add Equation 1 to Equation 3, then Equation 2 to Equation 4:
$\begin{cases}
A+C-A+2B-C-2D=0+0\\
A+B-C+D-A+B+C+D=1+0
\end{cases}$
$\begin{cases}
2B-2D=0\\
2B+2D=1
\end{cases}$
$2B-2D+2B+2D=0+1$
$4B=1$
$B=\dfrac{1}{4}$
$2B-2D=0$
$2\left(\dfrac{1}{4}\right)-2D=0$
$\dfrac{1}{2}=2D$
$D=\dfrac{1}{4}$
$A+C+A+B-C+D=0+1$\\
$2A+B+D=1$
$2A+\dfrac{1}{4}+\dfrac{1}{4}=1$
$2A=1-\dfrac{1}{4}-\dfrac{1}{4}$
$2A=\dfrac{1}{2}$
$A=\dfrac{1}{4}$
$A+C=0$
$\dfrac{1}{4}+C=0$
$C=-\dfrac{1}{4}$
The partial fraction decomposition is:
$\dfrac{x^2}{(x-1)^2(x+1)^2}=\dfrac{\dfrac{1}{4}}{x-1}+\dfrac{\dfrac{1}{4}}{(x-1)^2}-\dfrac{\dfrac{1}{4}}{x+1}+\dfrac{\dfrac{1}{4}}{(x+1)^2}$
$\dfrac{x^2}{(x-1)^2(x+1)^2}=\dfrac{1}{4(x-1)}+\dfrac{1}{4(x-1)^2}-\dfrac{1}{4(x+1)}+\dfrac{1}{4(x+1)^2}$
