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?

Here's an alternative way doing it. Rewrite the integral as:
$\int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}} dx = \int \frac{1 + \frac{x}{\sqrt{x^2-1}} dx}{x + \sqrt{x^2-1}}$.
Now use a simple substitution $u=x+\sqrt{x^2-1}$, so that the integral becomes:
$\int \frac{du}{u} = \log u + C = \log{x+\sqrt{x^2-1}}+C.$
