How do you integrate # 1/(1+e^x) # using partial fractions?

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Answer 1 · 2016-04-18T00:01:38

#int1/(1+e^x)dx = x-ln(1+e^x)+C#

We use substitution to put the problem into a form where we can use partial fractions.

Let #u = 1+e^x => du = e^xdx#. Then:

#int1/(1+e^x)dx = inte^x/(e^x(1+e^x)dx#

#=int1/(u(u-1)du#

Decomposing #1/(u(u-1))#, we find

#1/(u(u-1)) = 1/(u-1) - 1/u#

#=> int1/(u(u-1))du = int1/(u-1)du - int1/udu#

#=ln|u-1|-ln|u|+C#

#=ln|e^x+1-1|-ln|e^x+1|+C#

#=ln(e^x)-ln(e^x+1)+C#

#=x-ln(e^x+1)+C#

Rather than using partial fractions, we can also solve this with a little algebraic manipulation and substitution.

#int1/(1+e^x)dx = int(1+e^x-e^x)/(1+e^x)dx#

#=int(1+e^x)/(1+e^x)dx - int e^x/(1+e^x)dx#

#=intdx-inte^x/(1+e^x)dx#

#=x-inte^x/(1+e^x)dx#

Focusing on the remaining integral, let #u = (1+e^x) => du = e^xdx#

Substituting, we have

#inte^x/(1+e^x)dx = int1/udu#

#=ln|u|+C#

#=ln(1+e^x)+C#

Plugging this into the equation above, we can get our result:

#int1/(1+e^x)dx = x-ln(1+e^x)+C#