Evaluating $\int\frac{\sqrt{x^2-1}}x\mathrm dx$

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.

Put $X^2=x^2-1$ then $x^2=X^2+1$.

$2x\mathrm dx=2X\mathrm dX$.

$$\int\frac{\sqrt{x^2-1}}{x}\mathrm dx=\int\frac{X^2}{x^2}\mathrm dX=\int\frac{X^2}{X^2+1}\mathrm dX$$

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.)