91Ó°ÊÓ

The rational expression is,

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

Decomposition of the rational expression,

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

Multiplying both sides by ${(x-1)}^2{(x+1)}^2$,

role="math" localid="1647515334302" $$\begin{gathered} x^2=A(x-1){(x+1)}^2+B{(x+1)}^2+C(x+1){(x-1)}^2+D{(x-1)}^2{x}^2=A(x-1)(x^2+x+1)+B(x^2+x+1)+C(x+1)(x^2-2x+1)+D(x^2-2x+1) \\ {x}^2=A{x}^3+2A{x}^2+Ax-A{x}^2-2Ax-A+B{x}^2+2Bx+B+C{x}^3-2C{x}^2+Cx+C{x}^2-2Cx+C+D{x}^2-2Dx+D \\ {x}^2=(A+C)x^3(A+B-C+D)x^2+(-A+2B-C-2D)x+(-A+B+C+D)............\left(2\right) \end{gathered}$$

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

$$\begin{gathered} A+C=0.......\left(3\right) \\ A+B-C+D=1.......\left(4\right) \\ -A+2B-C-2D=0.........\left(5\right) \\ -A+B+C+D=0............\left(6\right) \end{gathered}$$

Solving the equations we get,

$$\begin{gathered} A=\frac{1}{4} \\ B=\frac{1}{4} \\ C=-\frac{1}{4} \\ D=\frac{1}{4} \end{gathered}$$

The partial fraction decomposition is,

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

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

Adding the rational expressions,

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

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

This solution is true.