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.

You had the "gist" of what you needed to do, but as others have noted, your substitution should yield the integrand $\dfrac{1}{u^2+1}$. 
We have
$$\int \frac {dx}{x \sqrt{x^2 - 1}} = \int \frac{x\,dx}{x^2\sqrt{x^2-1}}$$
As you did, we let $\, u=\sqrt{x^2-1}$. Then $du=\frac{x}{\sqrt{x^2-1}}\,dx$, so $x\,dx=\sqrt{x^2 - 1}\,du = u \,du$. 
Note that $$u = \sqrt{x^2 - 1} \implies u^2 = x^2 - 1 \iff x^2 = u^2 + 1 $$
So substituting gives us $$\int \frac{x\,dx}{x^2\sqrt{x^2-1}} = \int \dfrac{u \,du}{(u^2 + 1)u} = \int \frac {du}{u^2 + 1}$$
Now, we can use trigonometric substitution, and given a denominator of the form $u^2 + 1$, put $u = \tan \theta$. This gives us: $$\int \frac {du}{u^2 + 1} = \arctan(u) + C = \arctan(\sqrt{x^2 - 1}) + C$$
