How do you integrate #int e^-x/(1+e^-x)dx#?

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Answer 1 · 2018-03-10T13:32:35

# ln|1/(1+e^-x)|+C.#

Subst. #1+e^-x=t," so that, "-e^-xdx=dt#.

#:. I=inte^-x/(1+e^-x)dx#,

#=int-1/tdt=-int1/tdt#,

#=-ln|t|#,

#=ln|t|^-1#,

# rArr I=ln|1/(1+e^-x)|+C.#

Answer 2 · 2018-03-10T13:35:04

The answer is #=-ln(1+e^-x)+C#

Perform the substitution

#u=1+e^-x#, #=>#, #du=-e^-xdx#

Therefore,

#int(e^-xdx)/(1+e^-x)=-int(du)/(u)#

#=-lnu#

#=-ln(1+e^-x)+C#

Answer 3 · 2018-03-10T13:40:46

#I=int(e^-x)/(1+e^-x)dx=int((-1)d/dx(1+e^-x))/(1+e^-x)=-ln|1+e^-x|#
#I=ln|1/(1+e^-x)|+c#

#I=int(e^-x)/(1+e^-x)dx#,
take, #e^-x=t=>e^-x(-1)dx=dt=>e^-xdx=-dt#
#I=-intdt/(1+t)=-ln|1+t|+c=-ln|1+e^-x|+c#
#I=ln|1/(1+e^-x)|+c#