How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$ Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$


I attempted to solve it intergradation by parts by doing a 

$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$

and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$: 

but as I start doing, it gets more complicated. What is the right direction to solve this?
For $x^2-1$ in the radical use $x=\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$

For $x^2+1$ in the radical use $x=\tan\theta$  for the same reason

For $1-x^2$ in the radical use $x=\sin\theta$ as $1-\sin^2\theta=\cos^2\theta$
the solution is $= \operatorname{Argch(x)}$

because $\operatorname{Argch}^{\prime}(x)= \frac{1}{\sqrt{x^2-1}}$.