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