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)#