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How do I evaluate $$\int \frac{e^{-2x}}{1+e^{-x}} dx$$

I've tried the substitution u = $e^{-x}$, and the furthest I've gotten is $\int \frac{-u}{1+u} du$ and I don't know how to proceed from there.

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  • $\begingroup$ It's actually easier to substitute $u = 1 + e^{-x}$ $\endgroup$
    – Dylan
    Nov 23, 2014 at 19:28

2 Answers 2

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You've already done the hard part. Notice now that $$\int \frac{u}{1 + u} du = \int \left(1 - \frac{1}{1 + u}\right) du$$

$$\cdots = u - \log |u + 1| + C.$$

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  • $\begingroup$ That was exactly what I needed. Thanks! I'll accept your answer once the 10 mins limit is up. $\endgroup$
    – Vizuna
    Nov 16, 2014 at 15:27
  • $\begingroup$ You're welcome, I'm glad you found it helpful! $\endgroup$ Nov 16, 2014 at 15:28
  • $\begingroup$ I am puzzled by the nearly simultaneous downvotes on both my answer and Sami's very good answer... $\endgroup$ Nov 16, 2014 at 15:31
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    $\begingroup$ You're correct that, since $\log|x|$ is not connected, the general antiderivative of $\frac{1}{x}$ is the set of functions given by $\log|x| + C$ on $(-\infty, 0)$ and $\log|x| + D$ on $(0, \infty)$ for some constants $C, D$. But, nearly every calculus reference will suppress this complication, because in settings where the constant of integration is essential, one is already typically only working on one connected component of the domain, and good references justify this by declaring that expressions $+ C$ refer to working with one such component at a time, which is the sense I had in mind. $\endgroup$ Nov 16, 2014 at 16:23
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    $\begingroup$ @Joao Cheers! I was mostly just puzzled given that the question (and answer) are both so standard. Still, it is good to remind oneself from time to time of the sort of technical points that user2345215 mentions. $\endgroup$ Nov 17, 2014 at 3:23
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$$\int\frac{-u}{1+u}du=\int\frac{-u\color{red}{-1+1}}{1+u}du=-\int du+\int\frac{du}{1+u}=-u+\ln|1+u|+C$$

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