We have,
$x^{2}\left(x^{2}-1\right) \frac{d y}{d x}+x\left(x^{2}+1\right) y=x^{2}-1$
$\Rightarrow \quad \frac{d y}{d x}+\frac{x^{2}+1}{x\left(x^{2}-1\right)} y=\frac{1}{x^{2}}$
$\begin{aligned} I F=& e^{\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x} \\ &=e^{\int \frac{x^{2}+1}{x(x-1)(x+1)}} d x \end{aligned}$
Let $\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$
$\Rightarrow x^{2}+1=A(x-1)(x+1)+B x(x+1)$ $\quad+\operatorname{Cx}(x-1)$
$\begin{array}{ll}\text { Put } x=0, & \\ \therefore \quad & 1=-A \\ \Rightarrow & A=-1 \\ \Rightarrow &
\end{array}$
Put $x=1
\therefore \quad 2=2 B$
$\begin{array}{ll}\therefore & B=1 \\ \Rightarrow & x=-1\end{array}$
$\therefore \quad 2=2 C$
$\Rightarrow C=1$
$\therefore \frac{x^{2}+1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}$
$\therefore \quad I F=e^{\int\left(\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right) d x}$
$=e^{[-\log x+\ln (x-1)+\log (x+1)]}$
$=e^{\ln \left(\frac{x^{2}-1}{x}\right)}=\frac{x^{2}-1}{x}$
$=x-\frac{1}{x}$