91Ó°ÊÓ

The rational expression is,

$\frac{x^2}{{(x-1)}^2(x+1)}$.

Decomposition of the partial fraction,

$\frac{x^2}{{(x-1)}^2(x+1)}=\frac{A}{x-1}+\frac{B}{{(x-1)}^2}+\frac{C}{x+1}............\left(1\right)$

Multiplying both sides by,

$$\begin{gathered} x^2=A(x-1)(x+1)+B(x+1)+C{(x-1)}^2{x}^2=A(x^2-1)+B(x+1)+C(x^2+1-2x) \\ {x}^2=(A+C)x^2+(B-2C)x+(-A+B+C)...........\left(2\right) \end{gathered}$$

Equating the coefficients of the like power of $x$, we get,

$$\begin{gathered} A+C=1........\left(3\right) \\ B-2C=0......\left(4\right) \\ -A+B+C=0.......\left(5\right) \end{gathered}$$

$$\begin{gathered} B-2C=0 \\ ⇒B=2C.......\left(6\right) \end{gathered}$$

$$\begin{gathered} -A+B+C=0 \\ -A+2C+C=0 \\ -A+3C=0.........\left(7\right) \end{gathered}$$

$A+C=1.......\left(3\right)$

Solving the equations we get,

$$\begin{gathered} C=\frac{1}{4} \\ B=\frac{1}{2} \\ A=\frac{3}{4} \end{gathered}$$

The partial fraction decomposition is,

$\frac{x^2}{{(x-1)}^2(x+1)}=\frac{3}{4(x-1)}+\frac{1}{2{(x-1)}^2}+\frac{1}{4(x+1)}$.

Adding the rational expressions,

$\frac{3}{4(x-1)}+\frac{1}{2{(x-1)}^2}+\frac{1}{4(x+1)}=\frac{3(x-1)(x+1)+2(x+1)+{(x-1)}^2}{4{(x-1)}^2(x+1)}$

$$\begin{gathered} =\frac{3(x^2-1)+2(x+1)+(x^2+1-2x)}{4{(x-1)}^2(x+1)} \\ =\frac{3x^2-3+2x+2+x^2+1-2x}{4{(x-1)}^2(x+1)} \\ =\frac{4x^2}{4{(x-1)}^2(x+1)} \\ =\frac{x^2}{{(x-1)}^2(x+1)} \end{gathered}$$

The solution is true.