Mathematics

# Solve :$\displaystyle\int\dfrac {e^x}{-2(1+e^{-x} )^2} .dx$

##### SOLUTION
$\displaystyle I = \int\dfrac {e^x}{-2(1+e^{-x} )^2}dx$
$\displaystyle = \dfrac{-1}{2}\int \dfrac{e^{x}.e^{2x}}{(e^{x}+1)^{2}}dx$
put $e^{x} = t$
$e^{x}dx = dt$
$\displaystyle = -\dfrac{1}{2}\int \frac{t^{2}}{(t+1)^{2}}dt$
$\displaystyle = -\dfrac{1}{2}\int \frac{[(t+1)^{2}-2t-1]}{(t+1)^{2}}dt$
$\displaystyle I = [-\dfrac{1}{2}\int dt]+[\dfrac{1}{2}\int \dfrac{2t+1}{(t+1)^{2}}dt]$
$\displaystyle A = \int -\dfrac{1}{2}dt = \dfrac{-t}{2} = \dfrac{-e^{x}}{2}$
$A = \dfrac{-e^{x}}{2}$
$(B) \displaystyle \dfrac{1}{2}\int \dfrac{(2t+1)}{(t+1)^{2}}dt$
$\displaystyle = \dfrac{1}{2}\int \dfrac{2t+2}{(t+1)^{2}}dt-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt$
$\displaystyle \Rightarrow \dfrac{1}{2}\int \dfrac{\dfrac{d}{dx}(t+1)^{2}}{(t+1)^{2}}dt-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt$
$\displaystyle \Rightarrow \dfrac{1}{2}log |t+1|^{2}-\dfrac{1}{2}\int \dfrac{1}{(t+1)^{2}}dt$
$\displaystyle \Rightarrow log |P| -\dfrac{1}{2}\int \dfrac{dt}{(t+1)^{2}}$
$(t+1) = p$
$dt = dp$
$\displaystyle \Rightarrow log |p|-\frac{1}{2}\int \frac{dp}{p^{2}}$
$\displaystyle \Rightarrow log |p|-\frac{1}{2}(-\frac{1}{p})+c$
$\displaystyle \Rightarrow log |p|+\frac{1}{2p}+c$
$\displaystyle \Rightarrow B = log |t+1|+\frac{1}{2(t+1)}+c$
$\boxed{I = A+B = \dfrac{-e^{-x}}{2}+|log|e^{x}+1|+\dfrac{1}{2(e^{x}+1)}+c}$

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

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