How to integrate $\int{1\over \sqrt{x^2-1}}\mathrm d x$ another technique without use trigonometry

Question

How can I integrate $\int{1\over \sqrt{x^2-1}}\mathrm d x$ without using trygnometry? I mean using other methods like substituition, integration by parts (I tried these two but I think I am not seeing the best way to integrate. I reached some integrals more complicated.

Answer 1

Assume $x>1$ and write $${1\over\sqrt{x^2-1}}={1\over\sqrt{x^2-1}}\>{\sqrt{x^2-1}+x\over \sqrt{x^2-1}+x}={1+{x\over\sqrt{x^2-1}}\over x+\sqrt{x^2-1}}={u'(x)\over u(x)}$$ with $u(x):=x+\sqrt{x^2-1}>0$. It follows that $$\int{dx\over\sqrt{x^2-1}}=\log\bigl(u(x)\bigr)+C=\log\bigl(x+\sqrt{x^2-1}\bigr)+C\ .$$