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