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