Mathematics

# $\int { \frac { 1 }{ { e }^{ x }+1 } dx }$.

##### SOLUTION
$\displaystyle \int{\dfrac{1}{{e}^{x}+1}dx}$
$=\displaystyle \int{\dfrac{{e}^{-x}}{{e}^{-x}\left({e}^{x}+1\right)}dx}$
$=\displaystyle \int{\dfrac{{e}^{-x}}{1+{e}^{-x}}dx}$
Let $t={e}^{-x}+1\Rightarrow dt=-{e}^{-x}dx$
$=\displaystyle \int{\dfrac{-dt}{t}}$
$=-\ln{\left|t\right|}+c$ where $c$ is the constant of integration.
$=-\ln{\left|{e}^{-x}+1\right|}+c$
$=-\ln{\left|\dfrac{1}{{e}^{x}}+1\right|}+c$
$=-\left(\ln{\left|\dfrac{1+{e}^{x}}{{e}^{x}}\right|}\right)+c$
$=-\left(\ln{\left(1+{e}^{x}\right)}-\ln{{e}^{x}}\right)+c$
$=-\ln{\left(1+{e}^{x}\right)}+\ln{{e}^{x}}+c$
$=-\ln{\left(1+{e}^{x}\right)}+x+c$
$=x-\ln{\left(1+{e}^{x}\right)}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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