Evaluate the following integral:

$\frac{x^{2}}{x^{4}-x^{2}-12}=\frac{x^{2}}{\left(x^{2}-4\right)\left(x^{2}+3\right)}$

Let, $\frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}-4\right)\left(\mathrm{x}^{2}+3\right)}=\frac{\mathrm{A}}{\mathrm{x}-2}+\frac{\mathrm{B}}{\mathrm{x}+2}+\frac{\mathrm{C}}{\mathrm{x}^{2}+3}$

$\Rightarrow x^{2}=A(x+2)\left(x^{2}+3\right)+B(x-2)\left(x^{2}+3\right)+C(x-2)(x+2)$

For, $x=2, A=\frac{1}{7}$

For, $x=-2, B=-\frac{1}{7}$

For, $x=0, C=\frac{3}{7}$

$\therefore \int \frac{\mathrm{x}^{2}}{\mathrm{x}^{4}-\mathrm{x}^{2}-12} \mathrm{dx}=\frac{1}{7} \int \frac{\mathrm{dx}}{\mathrm{x}-2}-\frac{1}{7} \int \frac{\mathrm{dx}}{\mathrm{x}+2}+\frac{3}{7} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+3}$

$=\frac{1}{7} \log |x-2|-\frac{1}{7} \log |x+2|+\frac{3}{7 \sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+c$