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 February 4th, 2009, 05:59 AM #1 Newbie   Joined: Dec 2008 Posts: 26 Thanks: 0 Another Integral I faced another problem I can't solve by myself, and I need your help again. $\int_{}^{} \frac{dx}{e^{x} + e^{-x}}$ How to integrate this?
 February 4th, 2009, 07:36 AM #2 Member   Joined: Dec 2008 From: Washington state, of the United States Posts: 37 Thanks: 0 Re: Another Integral This is actually easier than it looks. $\int_{}^{} \frac{dx}{e^{x} + e^{-x}}$ Multiply through, top and bottom, by e^x $\int_{}^{} \frac{e^x}{e^{2x} + 1} dx$ Now, u-substitute (on the bottom) $u=e^x$ $\frac{du}{dx}= e^x$ $e^{-x} du= dx$ $\int_{}^{} \frac{e^x}{e^{2x} + 1} dx$ $\int_{}^{} \frac{e^x}{u^{2} + 1} dx$ $\int_{}^{} \frac{1}{u^{2} + 1} du$ This integrates easily. $\arctan(u) + C$ $\arctan(e^x) + C$
 February 4th, 2009, 08:28 AM #3 Newbie   Joined: Dec 2008 Posts: 26 Thanks: 0 Re: Another Integral Thanks again

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