# How to evaluate $\displaystyle{\int{\frac{1}{\sqrt{e^{2x}-4}}}\,dx}$

Please, could someone correct my exercise? I got it by myself, but the result that I found is differend by the WolframAlpha result... Could you tell me, please, if I'm wrong an where?

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Your answer is not correct. But even correct answers can look different, and WA often gives strange-looking answers. –  André Nicolas Jun 24 '12 at 17:13
The best way to check an antiderivative is to take its derivative and see if that simplifies to the integrand. –  Robert Israel Jun 24 '12 at 17:43

There is an error in the beginning. Setting $u^2 - 4 = s^2 \implies u^2 = s^2 +4$ and not $s^2-4$ as you have written. Because of this the entire integration is incorrect.
Put $e^{2x}-4=t^2$. :) –  B. S. Jun 24 '12 at 17:22
@BabakSorouh Got it after substitution with $u$ :P –  Overflowh Jun 24 '12 at 18:19
To solve the problem, it may be useful to let $e^x=2\sec t$. Then $e^x\,dx=2\sec t\tan t\,dt$, so $dx=\tan t\,dt$. The bottom turns out to be $2\tan t$, so we end up needing $$\int\frac{1}{2}\,dt.$$