Answer
The partial function is, $\frac{1}{4\left( x-1 \right)}+\frac{1}{4{{\left( x-1 \right)}^{2}}}-\frac{1}{4\left( x+1 \right)}+\frac{1}{4{{\left( x+1 \right)}^{2}}}$.
Work Step by Step
$\frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{{{\left( x-1 \right)}^{2}}}+\frac{C}{\left( x+1 \right)}+\frac{D}{{{\left( x+1 \right)}^{2}}}$.
Now, take L.C.M on right side:
$\frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}}=\frac{A\left( x-1 \right){{\left( x+1 \right)}^{2}}+B{{\left( x+1 \right)}^{2}}+C{{\left( x-1 \right)}^{2}}\left( x+1 \right)+D{{\left( x-1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}}$
By eliminating the denominators from both sides, we get:
$\begin{align}
& {{x}^{2}}=A\left( x-1 \right){{\left( x+1 \right)}^{2}}+B{{\left( x+1 \right)}^{2}}+C{{\left( x-1 \right)}^{2}}\left( x+1 \right)+D{{\left( x-1 \right)}^{2}} \\
& =A\left( x-1 \right)\left( {{x}^{2}}+1+2x \right)+B({{x}^{2}}+1+2x)+C\left( {{x}^{2}}+1-2x \right)\left( x+1 \right)+D\left( {{x}^{2}}+1-2x \right) \\
& =A\left( {{x}^{3}}+x+2{{x}^{2}}-{{x}^{2}}-1-2x \right)+B{{x}^{2}}+B+2Bx+C\left( {{x}^{3}}+{{x}^{2}}+x+1-2{{x}^{2}}-2x \right)+D{{x}^{2}}+D-2Dx \\
& =A\left( {{x}^{3}}+{{x}^{2}}-x-1 \right)+B{{x}^{2}}+B+2Bx+C\left( {{x}^{3}}-{{x}^{2}}-x+1 \right)+D{{x}^{2}}+D-2Dx
\end{align}$ That is
${{x}^{2}}=\left( A+C \right){{x}^{3}}+\left( A+B-C+D \right){{x}^{2}}+\left( -A+2B-C-2D \right)x+\left( -A+B+C+D \right)$
Then, compare the coefficient of ${{x}^{3}},{{x}^{2}},x$ and constant term:
$A+C=0$ (I)
$A+B-C+D=1$ (II)
$-A+2B-C-2D=0$ (III)
$-A+B+C+D=0$ (IV)
So, in the equation (III) put the value of equation (I):
$\begin{align}
& -\left( A+C \right)+2B-2D=0 \\
& 2B-2D=0 \\
& B-D=0
\end{align}$
$B=D$ (V)
And put equation (V) in equation (II):
$A-C+2B=1$ (VI)
And put equation (V) into equation (IV):
$-A+C+2B=0$ (VII)
And add equation (VI) and equation (VII):
$\begin{align}
& 4B=1 \\
& B=\frac{1}{4}=D
\end{align}$
Then, put the value of B in equation (VI):
$\begin{align}
& A-C+2\left( \frac{1}{4} \right)=1 \\
& A-C=1-\frac{1}{2}
\end{align}$
$A-C=\frac{1}{2}$ (VIII)
And add equation (VIII) with equation (I):
$\begin{align}
& 2A=\frac{1}{2} \\
& A=\frac{1}{4}
\end{align}$
Also, put the value of A in equation (I):
$\begin{align}
& \frac{1}{4}+C=0 \\
& C=\frac{-1}{4}
\end{align}$
Therefore,
$\frac{{{x}^{2}}}{{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}}=\frac{\frac{1}{4}}{\left( x-1 \right)}+\frac{\frac{1}{4}}{{{\left( x-1 \right)}^{2}}}+\frac{\frac{-1}{4}}{\left( x+1 \right)}+\frac{\frac{1}{4}}{{{\left( x+1 \right)}^{2}}}$.
Thus, the partial fraction of the provided expression is $\frac{1}{4\left( x-1 \right)}+\frac{1}{4{{\left( x-1 \right)}^{2}}}-\frac{1}{4\left( x+1 \right)}+\frac{1}{4{{\left( x+1 \right)}^{2}}}$.
