Question #5b977

1 Answer
Dec 16, 2017

#=ln(2)#

Explanation:

#int_0^∞1/(1+e^x)dx#

First, divide top and bottom by #e^x#

#=int_0^∞e^-x/(e^-x+1)dx#

Notice that the derivative of the denominator is almost the same as the current numerator, except its missing a #-# sign, so we fix that:
#=-int_0^∞-e^-x/(e^-x+1)dx#

Now we can do a u-substitution, letting #u = e^-x+1#, therefore #du = -e^-x dx# We also have to change the limits accordingly:
#x = 0 -> u = e^-0 + 1 = 2#
#x = ∞-> u = e^-∞ + 1 = 1#
Putting this in into the current integral,

#=-int_2^1 1/udu#
This is the well known #ln# integral
#=-[lnu]_2^1#
#=-ln1+ln2#
#=ln2#
#=ln(2)#