<s>[INST] Solving $\int \frac{1}{\sqrt{x^2 - c}} dx$ I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$
How do I do this?
It looks like it is close to being an $\operatorname{arcsin}$?

I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$
But apparently not. 
 [/INST] $$ \int \frac{1}{\sqrt{x^2-c}}dx = \int \frac{1}{\sqrt{c\left(\frac{x^2}{c}-1\right)}}dx=\frac{1}{\sqrt{c}} \int \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2-1}}dx $$
Let $u=\frac{x}{\sqrt{c}}$, then $du=\frac{1}{\sqrt{c}} dx$. So now we have
$$ \int \frac{1}{\sqrt{u^2-1}}du =\mathrm{arcosh}(u)+C = \mathrm{arcosh}\left( \frac{x}{\sqrt{c}}\right)+C $$
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