How to integrate ? $int_0^oo 1/(1+e^x)*dx$

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Answer 1 · 2017-05-06T00:09:17

$ln(2)$

First working with the indefinite integral, divide the numerator and denominator by $e^x$ :

$int1/(1+e^x)dx=inte^-x/(e^-x+1)dx$

Let $u=e^-x+1$ , implying that $du=-e^-xdx$ :

$=-int(-e^-x)/(e^-x+1)dx=-int1/udu=-lnabsu$

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

So:

$int_0^oo1/(1+e^x)dx$

$=(lim_(xrarroo)(-ln(e^-x+1)))-(-ln(e^0+1))$

Note that $lim_(xrarroo)e^-x=0$ :

$=-ln(1)+ln(1+1)$

$=ln(2)$

How to integrate ? int_0^oo 1/(1+e^x)*dxint_0^oo 1/(1+e^x)*dx