Question

How do you find `int(e^-x)/(1+e^-x)dx`?

Answer 1

`inte^(-x)/(e^(-x)+1)dx=x-ln(e^x+1)+c`

Explanation:

`inte^(-x)/(e^(-x)+1)dx=int(-(e^(-x)+1)')/(e^(-x)+1)dx=-int((e^(-x)+1)')/(e^(-x)+1)dx=-ln(e^(-x)+1)+c`

• `ln(e^(-x)+1)=ln(1/e^x+1)=ln((e^x+1)/e^x)=ln(e^x+1)-lne^x=ln(e^x+1)-x`

Therefore, `inte^(-x)/(e^(-x)+1)dx=-(ln(e^x+1)-x)+c=x-ln(e^x+1)+c`

, `c` `in` `RR`

Answer 2

The answer is `=-ln(1+e^-x)+C`

Explanation:

Perform this integral by substitution

Let `u=1+e^-x`

`=>`, `du=-e^-xdx`

Therefore, the integral is

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

`=-int(du)/u`

`=-ln(u)`

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

Answer 3

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

Explanation:

Let `u=e^{-x}`. This yields `du = -e^{-x}dx`

Rewrite

`\int \frac{e^{-x}}{1+e^{-x}}dx = - \int \frac{-e^{-x}}{1+e^{-x}}dx`

Substitute `-e^{-x}dx=du` and `e^{-x}=u`:

`- \int \frac{1}{1+u}du`

This integral is `-ln(1+u)+C`.

Substitute back `u=e^{-x}` to get

`\int \frac{e^{-x}}{1+e^{-x}}dx = -ln(1+e^{-x})+C`

Answer 4

I tried this:

Explanation:

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