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answer from book $x - \ln(e^x + 1)$ $~~~$ *I get * $- \ln \vert 1 + e^{-x} \vert + C $

$\int \frac{e^{-x}}{1+e^{-x}}\, \mathrm{d}x $

$\begin{array}{l c l} u & = & 1 + e^{-x} &\\ \mathrm{d}u & = & -e^{-x}\: \mathrm{d}x \end{array} $

$-\int \frac{\mathrm{d}u}{u} $

$\Rightarrow -\ln \vert 1 + e^{-x} \vert + C $

I don't think I made any mistake. How do I get the answer in the book?

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    Especially so when the integral lacks its $dx$.2012-11-21

2 Answers 2

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Just note that

$ \ln(e^{-x} + 1) = \ln(e^{-x}(1 + e^x)) = -x + \ln(1 + e^x). $

Also, note that you can drop the absolute values as $e^x$ (and hence $e^{-x}$) is always positive.

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Take $u=1+e^{-x}$. Then $-\text{du}=e^{-x}\text{dx}.$ Therefore, $\int \frac{e^{-x}}{1+e^{-x}}\text{dx}=-\int\frac{1}{u}\text{du}=-\ln|u|+C=-\ln|1+e^{-x}|+C=-\ln(1+e^{-x})+C,$ with the last equality following from the fact that $|1+e^{-x}|=1+e^{-x}$ since $1+e^{-x}>0.$