Evaluate the following integral:

let $x=t^{2}$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

$\int \frac{2 d t}{t^{4}+1}$

Dividing by $t^{2}$ in both numerator and denominator

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

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

Let $t-\frac{1}{t}=z a n d t+\frac{1}{t}=y$

$\left(1+\frac{1}{t^{2}}\right) d t=d z$ and $\left(1-\frac{1}{t^{2}}\right) d t=d y$

$\int \frac{d z}{z^{2}+2}-\int \frac{d y}{y^{2}-2}$

Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$ and $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$

$\frac{1}{\sqrt{2}} \arctan \left(\frac{z}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{y-\sqrt{2}}{y+\sqrt{2}}\right|+c$

Substituting $t-\frac{1}{t}=z$ and $t+\frac{1}{t}=y$

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

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