Mathematics

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

$\displaystyle x-\log \left ( e^{x} +1\right ).$

##### SOLUTION
Let $\displaystyle I=\int \frac { 1 }{ e^{ x }+1 } dx=\int \frac { e^{ x }dx }{ e^{ x }\left( e^{ x }+1 \right) }$
Put $\displaystyle e^{ x }=t\Rightarrow e^{ x }dx=dt$
$\displaystyle I=\int \frac { dt }{ t\left( t+1 \right) } =\int \left( \frac { 1 }{ t } -\frac { 1 }{ t+1 } \right) dt$
$\displaystyle =\log t-\log \left( t+1 \right) =\log e^{ x }-\log \left( e^{ x }+1 \right)$
$\displaystyle =x-\log \left( e^{ x }+1 \right)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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