Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$ 
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$

This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.

Integrate by parts
\begin{align}\int \sqrt{1 + \frac{1}{x^2}}\ dx
=x\sqrt{1 + \frac{1}{x^2}}+\int \frac{dx}{x^2 \sqrt{1 + \frac{1}{x^2}}}
=x\sqrt{1 + \frac{1}{x^2}}-\sinh^{-1}\frac1x
\end{align}
