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 #