Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral
$$\int \frac{dx}{x\sqrt{x^2-1}}$$
I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then
\begin{align}
&\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&\int \frac{1}{x} \text{d}u \\
&\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\ 
\end{align}
Now, this is where I am having trouble. How can I evaluate that? Please provide only hints
Thanks!
EDIT:
The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.

$$
\begin{aligned}\int \frac{d x}{x \sqrt{x^{2}-1}} =\int \frac{1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) =\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} =\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
