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}
&amp;\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&amp;\int \frac{1}{x} \text{d}u \\
&amp;\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.
$$
\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2\sqrt{x^2+1}} = \int\frac{1}{x^2\sqrt{x^2+1}}
\Big(x\,dx\Big)
$$

The big parentheses are of course a hint that what's inside them is to become $du$, or a constant times $du$.  But should $u$ be $x^2$ or $x^2+1$?  Either way, $\displaystyle\Big(x\,dx\Big)$ becomes $\displaystyle\Big( \frac12\,du\Big)$.  I think usually it's better to have the thing under the radical be simple, so I'll say $u=x^2+1$, and we have
$$
\frac12\int\frac{du}{(u-1)\sqrt{u}}.
$$
We can rationalize $\sqrt{u}$ by letting
\begin{align}
w &amp; = \sqrt{u} \\
w^2 &amp; = u \\
2w\,dw &amp; = du
\end{align}
and we have
$$
\frac12\int\frac{2w\,dw}{(w^2-1)w} = \int\frac{dw}{w^2-1}.
$$
Then use partial fractions, getting
$$
\int\left(\frac{A}{w-1}+\frac{B}{w+1}\right)\,dw
$$
and you need to figure out what $A$ and $B$ are.

That works, but a trigonometric substitution also comes to mind.  The expression $\sqrt{x^2+1}$ should remind you of $\sqrt{\tan^2\theta+1}= \pm\sec\theta$, and if it doesn't remind you of that, that's something to work on.  Review some trigonometry and trigonometric substitutions.  If $x=\tan\theta$ then $dx=\sec^2\theta\,d\theta$, and we have
$$
\int\frac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta} = \int\frac{\sec\theta\,d\theta}{\tan\theta} = \int\csc\theta\,d\theta.
$$
That's a hard one to do from scratch, but it's also one that you can look up in standard tables.
When $x =\sin u$, $\displaystyle\int\frac{dx}{x\sqrt{x^2 -1}} \, dx$ becomes $\displaystyle\int\frac{\cos u \, du}{\sin u\sqrt{-\cos^2u}}$.