# Math Help - Unusual Integral

1. ## Unusual Integral

Evaluate

$\int \frac{1}{e^x + 1} dx$.

2. $-ln(e^x+1)+x$

3. $\frac{1}{e^x + 1} \times \frac{e^{-x}}{e^{-x}}= \frac{e^{-x}}{1 + e^{-x}}$

Can you go from here?

4. Originally Posted by icemanfan
Evaluate

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

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

$= x - \ln(e^x + 1) + c$

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

Correct?

6. Originally Posted by icemanfan
$\int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C$

Correct?
You lost an x and your e should be to the positive power.

7. Originally Posted by dwsmith
You lost an x and your e should be to the positive power.
Nah it's right I think.

Someone show me how the two answers we have given are equiv though..?

8. Originally Posted by Deadstar
Nah it's right I think.

Someone show me how the two answers we have given are equiv though..?
Yeah they are equivalent. I checked them in Maple; however, I have no idea how to show that they are.

9. Originally Posted by Deadstar
Nah it's right I think.

Someone show me how the two answers we have given are equiv though..?
-ln(1-e^-x) = -ln(1 - 1/e^x) = -ln[(e^x - 1)/e^x]
= -[ln(e^x - 1) - lne^x]
= x- ln(e^x - 1)