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.

$$u^{2}=x^{2}-1\Rightarrow xdx=udu\\
\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{xdx}{(x^{2}-1+1)\sqrt{x^2-1}}\\
=\int\frac{udu}{(u^{2}+1)u} =\int\frac{du}{u^{2}+1}=\arctan(u)+c=\arctan\sqrt{x^{2}-1}+c$$
