Evaluate the following integral:

re-writing the given equation as

$\int \frac{\frac{1}{x^{2}}}{x^{2}+3+\frac{1}{x^{2}}} d x$

$\frac{1}{2} \int \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+3+\frac{1}{x^{2}}} d x$

$\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+5} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}+1}\right]$

Assume $t=x-\frac{1}{x}$ and $z=x+\frac{1}{x}$

$\mathrm{dt}=\left(1+\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}$ and $\mathrm{dz}=\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}$

$\frac{1}{2}\left[\int \frac{\mathrm{dt}}{(\mathrm{t})^{2}+5}-\int \frac{\mathrm{dz}}{(\mathrm{z})^{2}+1}\right]$

Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$

$\frac{1}{2 \sqrt{5}} \arctan \left(\frac{\mathrm{t}}{\sqrt{5}}\right)-\frac{1}{2} \arctan (\mathrm{z})+\mathrm{c}$

Substituting $t$ as $x-\frac{1}{x}$ and $z$ as $x+\frac{1}{x}$

$\frac{1}{2 \sqrt{5}} \arctan \left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2} \arctan \left(x+\frac{1}{x}\right)+c$