How do you integrate # (1/(e^x+1))dx #?
2 Answers
Explanation:
Let
From
Explanation:
Note that:
#1/(e^x+1) = e^(-x)/(1+e^(-x)) = -d/(dx) ln (1+e^(-x))#
So:
#int \ 1/(e^x+1) \ dx = -ln(1+e^(-x))+C#
If you prefer, note that:
#-ln(1+e^(-x)) = -ln((e^x+1)/(e^x))#
#color(white)(-ln(1+e^(-x))) = ln(e^x)-ln(e^x+1)#
#color(white)(-ln(1+e^(-x))) = x-ln(e^x+1)#
So the integral can be expressed as:
#int \ 1/(e^x+1) \ dx = x-ln(e^x+1)+C#