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)