Evaluate the following integral:

Let, I $=\int \frac{x^{2}}{x^{4}+x^{2}-2} d x$

Let, $\frac{\mathrm{x}^{2}}{\mathrm{x}^{4}+\mathrm{x}^{2}-2}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}-1}+\frac{\mathrm{C}}{\mathrm{x}^{2}+2}$

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

For, $x=1, A=\frac{1}{6}$

For, $x=-1, B=-\frac{1}{6}$

For, $x=0, C=-\frac{2}{3}$

$\therefore \mathrm{I}=\frac{1}{6} \int \frac{\mathrm{dx}}{\mathrm{x}+1}-\frac{1}{6} \int \frac{\mathrm{dx}}{\mathrm{x}-1}-\frac{2}{3} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+2}$

$\Rightarrow I=\frac{1}{6} \log |x+1|-\frac{1}{6} \log |x-1|-\frac{2}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+c$