Evaluate the integral int \ 1/(1+e^x) \ dx ?

1 Answer
Dec 13, 2017

int \ 1/(1+e^x) \ dx = -ln |1+e^(-x)| + C

Explanation:

We seek:

I = int \ 1/(1+e^x) \ dx
\ \ = int \ 1/(1+e^x) \ e^(-x)/e^(-x) \ dx
\ \ = int \ e^(-x)/(1+e^(-x)) \ dx

Perform the substitution:

u = 1+e^(-x) => (du)/dx = -e^(-x)

Then the integral becomes:

I = int \ 1/(u) (-1) \ du
\ \ = - \ int \ 1/(u) \ du

Which is now a trivial integral, so integrating we get:

I = -ln |u| + C

And restoring the substitution:

I = -ln |1+e^(-x)| + C