<s>[INST] 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.
 [/INST] 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: $$\int \frac {du}{u^2 + 1} = \arctan(u) + C = \arctan(\sqrt{x^2 - 1}) + C$$
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