Question

How do you integrate `(1/(e^x+1))dx`?

Answer 1

`x-ln(e^x+1)+C`

Explanation:

Let `e^(x/2)=tantheta`. Then `1/2e^(x/2)dx=sec^2thetad theta`.

`intdx/(e^x+1)=2int(1/2e^(x/2)dx)/(e^(x/2)(e^x+1))=2int(sec^2thetad theta)/(tantheta(sec^2theta))=2intcostheta/sinthetad theta`

`=2lnabssintheta`

From `tantheta=e^(x/2)` draw a right triangle to see that `sintheta=e^(x/2)/sqrt(e^x+1)`:

`=2lnabs(e^(x/2)/sqrt(e^x+1))=lnabs(e^x/(e^x+1))=x-ln(e^x+1)+C`

Answer 2

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

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`