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?
 A: Hint let $u=\tan (y) $ and $dx=\sec^2 (y)dy$ and then continue to get the integral which can be calculated
A: \begin{align}
x & = \cosh u \\
dx & = \sinh u \, du \\[10pt]
\int \frac{2x^2}{\sqrt{x^2-1}}\, dx
&=\int \dfrac{2\cosh^2 u}{\sqrt{\cosh^2u-1}}\, \sinh (u) \, du \\[10pt]
& =\int {2\cosh^2 u} \, du \\[10pt]
& =\int {(\cosh (2u) + 1) } \, du \\[10pt]
& = (1/2) \sinh 2u + u \\[10pt]
& = (1/2) 2\cosh u\sinh u + u \\[10pt]
& = (1/2) \cosh u \sqrt {\cosh^2 u - 1} + u \\[10pt]
& = x \sqrt {x^2 - 1} + \cosh^{-1}(x) + k
\end{align}
$\operatorname{arccosh}$ can be expressed in terms of $\ln$ - note I developed some identities for $\cosh^2$ and $\cosh2x$, they can easily be found on line also, probably
A: my hint:
$$\int \frac{2x^2}{\sqrt{x^2-1}}\ dx=\int \frac{2(x^2-1)+2}{\sqrt{x^2-1}}\ dx=2\int\sqrt{x^2-1}\ dx+2\int\frac{1}{\sqrt{x^2-1}}\ dx$$
both integrals can be evaluated by substituting $u=\sec \theta$
A: \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.
