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.
On
set $t=\sqrt{x^2-1}$ then we get $$dx=\frac{\sqrt{x^2-1}}{x}dt$$ and we get $$\int\frac{4dt}{t^2+1}$$
On
We have, $$\int \frac{4}{x\sqrt{x^2-1}}dx$$ Let, $x=sec\alpha\implies dx=\sec\alpha\tan \alpha d\alpha $ $$=\int \frac{4\sec\alpha\tan \alpha d\alpha}{\sec\alpha\sqrt{\sec^2\alpha-1}}dx $$ $$=4\int \frac{\tan \alpha d\alpha}{\tan\alpha}dx =4\int d\alpha=4\alpha+C=4\sec^{-1}(x)+C$$
On
If $x\ge 1$, set $x=\cosh u,\enspace u\ge0$. Then you have: \begin{align*}\int\frac{4\sinh u\,\mathrm d\mkern1mu u}{\cosh u\sqrt{\cosh^2u-1 }}=4\int\frac{\mathrm d\mkern1mu u}{\cosh u}=8\int\frac{\mathrm e^u\,\mathrm d\mkern1mu u}{\mathrm e^{2u}+1}=8\arctan\mathrm e^u=8\arctan(x+\sqrt{x^2-1}). \end{align*} If $x\le-1$, setting $x=-\cosh u,\enspace u\ge 0$, we obtain the same integral, hence $$x=-8\arctan(x+\sqrt{x^2-1}).$$
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 $$4\int\frac{1}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{x^{2}}\frac{x}{\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du. $$.