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

1 Answer
May 6, 2017

ln(2)

Explanation:

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)