<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] 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 $u=\sec \theta$
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