Question

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

Answer 1

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

Explanation:

Rather than using partial fractions directly we can prepare the function for a substitution. As the substitution will have to remove the exponential, let's bring it to the numerator noting that: `1/e^x = e^-x`:

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

Substitute now `e^-x = t`, `dt =-e^-xdx`

`int dx/(e^x(e^x+1)) = - int (dt)/(1/t+1) = - int(tdt)/(t+1)`

Now separate in partial fractions simply adding and subtracting `1` to the numerator:

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

This are regular integrals we can solve straight away:

`int dx/(e^x(e^x+1)) = -t+ln abs (t+1) +C`

and undoing the substitution:

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