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