TITLE: Evaluating the following integral: $ \int \frac{x^2}{\sqrt{x^2 - 1}} \text{ d}x$
QUESTION [5 upvotes]: For this indefinite integral, I decided to use the substitution $x = \cosh u$ and I've ended up with a $| \sinh u |$ term in the denominator which I'm unsure about dealing with: $$\int \dfrac{x^2}{\sqrt{x^2 - 1}} \text{ d}x \ \overset{x = \cosh u}= \int \dfrac{\cosh^2 u \cdot \sinh u}{\left| \sinh u \right|} \text{ d}u$$
How would I deal with the denominator?

REPLY [2 votes]: First assume that $x=\tan u$, after replacing this value, $dx=du/(\cos u)^2$
in terms of $u$ variable, the integral will be $$I=\int\cos u (\sin u)^2 du.$$
Then replace $sin u= v$, you gotta find that $\cos u du=dv$. With v variable, we will have $I=\int v^2 dv$
and the result (answer) in terms of $v$ variable is $\tfrac{v^3}3+c$
going back to $u$ variable, we have this solution $\tfrac{\sin^3 u}3+c$
and by replacing $u$ by its value $\arctan x$, the result is: $$\boxed{\displaystyle\dfrac13 \sin ^3(\arctan x)+c}$$