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

Question

19706 views around the world

You can reuse this answer
Creative Commons License

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

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