Originally Posted by

**Jhevon** you can make this work (i would have probably tried a trig substitution, but your way is fine, probably easier).

$\displaystyle \int \frac 1{x \sqrt{x^2 - 1}}~dx$

note that this is the same as: $\displaystyle \int \frac x{x^2 \sqrt{x^2 - 1}}~dx$ you'll see why i did that in a minute.

Let $\displaystyle u = \sqrt{x^2 - 1} \implies x^2 = u^2 + 1$

$\displaystyle \Rightarrow du = \frac {x}{\sqrt{x^2 - 1}}~dx$

So, our integral becomes

$\displaystyle \int \frac 1{\left( u^2 + 1 \right)u}~du$

by partial fractions we find that this is:

$\displaystyle \int \left(\frac 1u - \frac u{u^2 + 1} \right)~du$

and i think you can take it from there, right?