Evaluate the following integral:

assume $\mathrm{X}=\frac{1}{\mathrm{t}}$

$\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$

Let $1+t^{2}=u^{2}$

$\mathrm{tdt}=\mathrm{udu}$

$\int \frac{u d u}{\left(u^{2}-2\right) u}$

$\int \frac{d u}{\left(u^{2}-2\right)}$

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

$\frac{1}{2 \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+c$

Substituting $u=\sqrt{1+t^{2}}$

$\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{1+t^{2}}-\sqrt{2}}{\sqrt{1+t^{2}}+\sqrt{2}}\right|+c$

Substituting $t=\frac{1}{x}$

$\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{1+\frac{1}{x^{2}}}-\sqrt{2}}{\sqrt{1+\frac{1}{x^{2}}}+\sqrt{2}}\right|+c$