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.