Evaluating $\int\sqrt{1 + \frac1{x^2}}\mathrm dx$ How do I solve this integral?
$$\int \sqrt{1 + \frac1{x^2}}\mathrm dx$$
I tried substituting $x=\cot t$ but ended up getting another tricky integral, so any help would be appreciated.

Hint :$$I=\int \frac{x}{\sqrt{x^2+1}}dx+\int \frac{1}{x\sqrt{x^2+1}}dx,\\
\int \frac{1}{x\sqrt{x^2+1}}dx=-\int\frac{d(\frac{1}{x})}{\sqrt{1+\frac{1}{x^2}}}$$
