Question

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

Answer 1

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

Explanation:

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`

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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`