Evaluating $\int\frac{\sqrt{x^2-1}}x\mathrm dx$ How can one evaluate the integral
$$\int\frac{\sqrt{x^2-1}}x\mathrm dx$$?
I tried substituting $x = \cosh t$ but got stuck at
$$\int\frac{\sinh^2t}{\cosh t}\mathrm dt$$
Any hints?
 A: I'll use the hyperbolic substitution you made.  (Why not?)  Of importance is the hyperbolic dual of the Pythagorean identity, $\cosh^2 x - \sinh^2 x = 1$.  Then, one can see that:
$$\frac{\sinh^2 t}{\cosh t} = \frac{\cosh^2 t - 1}{\cosh t} $$
This makes your integral:
$$\int \cosh t - \operatorname{sech} t\,dt$$
If you know your hyperbolic trig integrals as well as most people know their "normal" trig integrals, you're home free.
Hint:
$$\int\operatorname{sech} t\,dt = 2\arctan\left(\tanh\left(\frac{t}{2}\right)\right) + C$$
(According to Wolfram.)
A: Let $ x = \sec u $. Then, $ \mathrm{d}x = \sec u \tan u \, \mathrm{d}u $. Then, the integral becomes $$ \int \tan^2 u \, \mathrm{d}u = \int \left( \sec^2 u - 1 \right) \, \mathrm{d}u = \tan u - u + \mathcal{C}. $$ Then, you can substitute back and finish. 
A: Put $X^2=x^2-1$ then $x^2=X^2+1$.
$2x\mathrm dx=2X\mathrm dX$.
$$\int\frac{\sqrt{x^2-1}}{x}\mathrm dx=\int\frac{X^2}{x^2}\mathrm dX=\int\frac{X^2}{X^2+1}\mathrm dX$$
A: $$
\begin{aligned}\int \frac{\sqrt{x^{2}-1}}{x} d x =& \int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
=& \int \frac{x^{2}-1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) \\
=& \int\left(1-\frac{1}{x^{2}}\right) d\left(\sqrt{x^{2}-1}\right) \\
=& \sqrt{x^{2}-1}-\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} \\
=& \sqrt{x^{2}-1}-\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
