Question

Question #5b977

Answer 1

`=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)`