How do you evaluate the integral #int e^t/(e^t+1)dt#?

1 Answer
May 6, 2017

# int \ (e^t)/(e^t+1) \ dt = ln(e^t+1)+C #

Explanation:

We want to find:

# I = int \ (e^t)/(e^t+1) \ dt #

We can perform a simple substitution; Let

# u = e^t+1 => (du)/dt = e^t #

If we perform the substitution then we get:

#I=int color(red)(e^t)/u " " (du)/color(red)(e^t)#
# \ \ = int \ 1/u \ du #
# \ \ = ln|u| + C #

And restoring the substitution we get:

# I = ln(e^t+1)+C #