Question

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

Answer 1

$I=\int \frac{e^x}{-2(1+e^{-x}{)}^2}dx$

$=\frac{-1}{2}\int \frac{e^x.e^{2x}}{(e^x+1{)}^2}dx$

put $e^x=t$

$e^{x}dx=dt$

$=-\frac{1}{2}\int \frac{t^2}{(t+1{)}^2}dt$

$=-\frac{1}{2}\int \frac{\left[\right(t+1{)}^2-2t-1]}{(t+1{)}^2}dt$

$I=[-\frac{1}{2}\int dt]+[\frac{1}{2}\int \frac{2t+1}{(t+1{)}^2}dt]$

$A=\int -\frac{1}{2}dt=\frac{-t}{2}=\frac{-e^x}{2}$

$A=\frac{-e^x}{2}$

$\left(B\right)\frac{1}{2}\int \frac{(2t+1)}{(t+1{)}^2}dt$

$=\frac{1}{2}\int \frac{2t+2}{(t+1{)}^2}dt-\frac{1}{2}\int \frac{1}{(t+1{)}^2}dt$

$\Rightarrow \frac{1}{2}\int \frac{\frac{d}{dx}(t+1{)}^2}{(t+1{)}^2}dt-\frac{1}{2}\int \frac{1}{(t+1{)}^2}dt$

$\Rightarrow \frac{1}{2}log|t+1{|}^2-\frac{1}{2}\int \frac{1}{(t+1{)}^2}dt$

$\Rightarrow log|P|-\frac{1}{2}\int \frac{dt}{(t+1{)}^2}$

$(t+1)=p$

$dt=dp$

$\Rightarrow log|p|-\frac{1}{2}\int \frac{dp}{p^2}$

$\Rightarrow log|p|-\frac{1}{2}(-\frac{1}{p})+c$

$\Rightarrow log|p|+\frac{1}{2p}+c$

$\Rightarrow B=log|t+1|+\frac{1}{2(t+1)}+c$

$I=A+B=\frac{-e^{-x}}{2}+|log|e^x+1|+\frac{1}{2(e^x+1)}+c$