# Solving integral $\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx$

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. :) )

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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 $$\int \frac{\sqrt{1 - x^2} - 1}{x^2 - 1}dx = -\arcsin(x) + \text{atanh}(x) +C.$$

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Or you can just use partial fractions on the second term to avoid arc hyperbolic tangent. –  Graphth Apr 11 '11 at 19:38
If you try to avoid an inverse hyperbolic tangent, you end up with a mess of logarithms that can be combined to form the inverse hyperbolic tangent. So, what's the point? –  Guess who it is. Apr 12 '11 at 5:58

Replace $\cos^2(u)$ in the denominator by $1-\sin^2(u)$ and use substitution.

Or simpler, split the first integral in two simpler ones.

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