TITLE: Solving integral $\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx$
QUESTION [5 upvotes]: I've been asking a lot of integral questions lately. :D This is the integral I'm trying to solve: $$\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx$$
By replacing $x = \sin(u)$ (thus $dx = \cos(u)du$ and $u = \arcsin(x)$) I arrived at: $$\int \frac{\cos(u)}{\cos^2(u)}du - u + C$$
That fraction I think is $\sec(u)$, but we never learned about the secant function in school so I'd rather not use that. (Doesn't mean I don't want to know how to use it, I just want to be able to solve this some other way. :) )

REPLY [5 votes]: Why do you propose this substitution?
Your original integral can be split into two parts, $$
\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx = 
-\int \left(\frac{1}{\sqrt{1 - x^2}} + \frac{1}{x^2 - 1} \right)dx .$$
The antiderivative of the first term is given by $\arcsin(x)$, the antiderivate of the second term is given by $-\text{atanh}(x)$. So the total antiderivative is given by
$$\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx = -\arcsin(x) + \text{atanh}(x) +C.$$