91Ó°ÊÓ

The given rational expression is

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

Partial fraction decomposition of rational expression

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x^2}{\left(x^2+1\right)(x^2-1)}=\frac{Ax+B}{\left(x^2+1\right)}+\frac{C}{\left(x+1\right)}+\frac{D}{(x-1)} \\ \frac{x^2}{\left(x^2+1\right)(x^2-1)}=\frac{\left(Ax+B\right)\left(x^2-1\right)}{\left(x^2+1\right)(x^2-1)}+\frac{C\left(x^2+1\right)\left(x-1\right)}{\left(x^2+1\right)(x^2-1)}+\frac{D\left(x^2+1\right)\left(x+1\right)}{\left(x^2+1\right)(x^2-1)} \end{gathered}$$

Rearrange the equation

$$\begin{gathered} \frac{x^2}{\left(x^2+1\right)(x^2-1)}=\frac{\left(Ax+B\right)\left(x^2-1\right)}{\left(x^2+1\right)(x^2-1)}+\frac{C\left(x^2+1\right)\left(x-1\right)}{\left(x^2+1\right)(x^2-1)}+\frac{D\left(x^2+1\right)\left(x+1\right)}{\left(x^2+1\right)(x^2-1)} \\ {x}^2=\left(Ax+B\right)\left(x^2-1\right)+C\left(x^2+1\right)\left(x-1\right)+D\left(x^2+1\right)\left(x+1\right) \\ \left(0\right)x^3+\left(1\right)x^2+\left(0\right)x+0=(A+C+D)x^3+(B-C+D)x^2+(C+D-A)x+\left(D-B-C\right) \end{gathered}$$

Equate the constant and coefficients

$$\begin{gathered} A+C+D=0⋯i \\ B-C+D=1⋯ii \\ -A+C+D=0⋯iii \\ -B-C+D=0⋯iv \end{gathered}$$

Subtract equation iv from ii

$$\begin{gathered} \left(B-C+D\right)-(-B-C+D)=1-0 \\ B=\frac{1}{2} \end{gathered}$$

Subtract equation iii from i

$$\begin{gathered} \left(A+C+D\right)-\left(-A+C+D\right)=0-0 \\ A=0 \end{gathered}$$

Subtract equation ii from i and substitute values of A and B

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

Substitute value of A and C in equation i

$$\begin{gathered} 0-\frac{1}{4}+D=0 \\ D=\frac{1}{4} \end{gathered}$$

So partial fraction decomposition is

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