Question

How do you use partial fraction decomposition to decompose the fraction to integrate `1/(1+e^x)`?

Answer 1

`int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C`.

Explanation:

This is not really a partial fraction problem, but instead it's just a trick. Write:

`1/(1+e^(x))=(1+e^(x)-e^(x))/(1+e^(x))`

`=(1+e^(x))/(1+e^(x))-e^(x)/(1+e^(x))=1-e^(x)/(1+e^(x))`

This means

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

`=x-int e^(x)/(1+e^(x))\ dx`.

For this last integral, let `u=1+e^(x)` so that `du=e^(x)\ dx` and we get

`int e^(x)/(1+e^(x))\ dx=int\ (du)/u=ln|u|+C=ln(1+e^(x))+C` (note that `1+e^(x)>0` for all `x`).

Therefore,

`int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C`.