<s>[INST] Integration of $\frac{2x^2}{\sqrt{x^2-1}}\, dx$ Integration of $\dfrac{2x^2}{\sqrt{x^2-1}}\,\mathrm dx$
My try:
Let $u=\sqrt{x^2-1}$
Then $x=\sqrt{u^2+1}$
$x \,\mathrm dx =u\,\mathrm  du$
$$\int \frac{2u(u^2+1)}{u\sqrt{u^2+1}} \, \mathrm du$$
$$=2\int \sqrt {u^2+1} \, \mathrm du$$
True ? and what about the last integration?
 [/INST] \begin{align}
x & = \sec\theta \\[5pt]
dx & = \sec\theta\tan\theta\, d\theta \\[5pt]
\sqrt{x^2-1} & = \tan\theta \\[5pt]
\frac{2x^2}{\sqrt{x^2-1}}\, dx & = \frac{2\sec^2\theta}{\tan\theta} \sec\theta\tan\theta\,d\theta = 2\sec^3\theta\,d\theta
\end{align}
The integral of secant cubed is well known to be challenging to those first encountering it, but it does not require advanced methods.
</s>