How do you evaluate the integral of $int e^-x/(1+e^-2x) dx$ ?

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Answer 1 · 2016-06-07T12:11:17

$int (e^(-x)dx)/(1+e^(-2x))=-arctan e^(-x)+C$

$int (e^(-x)dx)/(1+e^(-2x))=-tan^(-1) e^(-x)+C$

$int e^(-x)/(1+e^(-2x))dx$

$int e^(-x)/(1^2+(e^(-x))^2)dx$

$-int (e^(-x)(-1)*dx)/(1^2+(e^(-x))^2)$

We can now use the formula

$int (du)/(u^2+a^2)=1/a*arctan (u/a)+C$

$int e^(-x)/(1+e^(-2x))dx=-int (e^(-x)(-1)*dx)/(1^2+(e^(-x))^2)$

$=(-1/1)*arctan(e^(-x)/1)+C$

$=-arctan e^(-x)+C$

$=-tan^(-1) e^(-x)+C$

God bless....I hope the explanation is useful.

How do you evaluate the integral of int e^-x/(1+e^-2x) dxint e^-x/(1+e^-2x) dx?