# Unusual Integral

Printable View

• April 12th 2010, 04:55 PM
icemanfan
Unusual Integral
Evaluate

$\int \frac{1}{e^x + 1} dx$.
• April 12th 2010, 05:03 PM
dwsmith
$-ln(e^x+1)+x$
• April 12th 2010, 05:04 PM
pickslides
$\frac{1}{e^x + 1} \times \frac{e^{-x}}{e^{-x}}= \frac{e^{-x}}{1 + e^{-x}}$

Can you go from here?
• April 12th 2010, 05:06 PM
Deadstar
Quote:

Originally Posted by icemanfan
Evaluate

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

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

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

$= x - \ln(e^x + 1) + c$
• April 12th 2010, 05:12 PM
icemanfan
$\int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C$

Correct?
• April 12th 2010, 05:15 PM
dwsmith
Quote:

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.
• April 12th 2010, 05:17 PM
Deadstar
Quote:

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..?
• April 12th 2010, 05:38 PM
dwsmith
Quote:

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.
• April 12th 2010, 05:53 PM
sa-ri-ga-ma
Quote:

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)