Mathematics

# Evaluate $\int {\frac {e^{2x}}{1+e^x}}dx$

##### SOLUTION
$\displaystyle l = \int \dfrac{e^{2x}}{1+e^{x}}dx$
$\displaystyle = \int \dfrac{e^{x}e^{x}}{1+e^{x}}dx$
Let $(1+e^{x}) = t$
$e^{x} = (t-1)$
$e^{x}dx = dt$
$\displaystyle l = \int \dfrac{(t-1)}{t}dt$
$\displaystyle \int (1-\dfrac{1}{t})dt$
$= (t-logt)+c$
$= e^{x}-loge^{x}+c$
$= e^{x}-x+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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