<s>[INST] How to integrate $\int \frac{4}{x\sqrt{x^2-1}}dx$ In order to solve the following integral:
$$\int \frac{4}{x\sqrt{x^2-1}}dx$$
I tried different things such as getting $u = x^2 + 1$, $u=x^2$ but it seems that it does not work. I also tried moving $x$ to the radical and writing the denominator as $\sqrt{x^4-x^2}$ but it also didn't work. Its look reminds me of $$\int \frac {1}{\sqrt{a^2-x^2}}dx = \arcsin\left(\frac{x}{a}\right) +c$$ but there is an extra $x$ in the denominator that I can't get rid of.
 [/INST] We have, taking $u=\sqrt{x^{2}-1}
 $, $$\int\frac{4}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.
 $$ I think you can get from here. Expanding a little bit, note that $$ u=\sqrt{x^{2}-1}\Rightarrow du=\frac{x}{\sqrt{x^{2}-1}}dx
 $$ and $$ x^{2}=u^{2}+1
 $$ then $$4\int\frac{1}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{x^{2}}\frac{x}{\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.
 $$.
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