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In fact, if you multiply the numberator and the denominator by <math>e^{-x}</math>, the integral becomes | In fact, if you multiply the numberator and the denominator by <math>e^{-x}</math>, the integral becomes | ||
| − | <math>\int\frac{1}{1+e^{-x}} e^{-x}\ dx =\int-\frac{1}{u}\ du,</math> | + | <math>\int\frac{1}{1+e^{-x}}\ e^{-x}\ dx =\int-\frac{1}{u}\ du,</math> |
where <math>u=1+e^{-x}</math>. | where <math>u=1+e^{-x}</math>. | ||
Revision as of 06:46, 1 October 2008
Evaluate the integral: $ \int \frac{dx}{1+e^x} $
I tried setting 'u' equal to '1+e^x' and 'du' equal to 'e^x dx'. Anyone know where to go from here? I'm sure there's some kind of manipulation that needs to be pulled off. I'll tell you if I figure it out.Gbrizend
Alright, I factored out a $ e^x $. I set 'u' equal to $ 1+\frac{1}{e^x} $ and 'du' to $ -1\frac{1}{e^x} dx $. The problem looks like this: $ \int \frac{dx}{e^x(1+\frac{1}{e^x})} $. Substitute with u and du. I got this answer: $ -Ln|1+\frac{1}{e^x}| + C. $
Good work. That last integral is easier to look at if you write $ e^{-x} $ in place of $ \frac{1}{e^x} $. In fact, if you multiply the numberator and the denominator by $ e^{-x} $, the integral becomes
$ \int\frac{1}{1+e^{-x}}\ e^{-x}\ dx =\int-\frac{1}{u}\ du, $
where $ u=1+e^{-x} $.
Here's another way. Let $ u=1+e^x $. Then $ du=e^x dx $ and $ e^x=u-1 $. If I multiply and divide by $ e^x $ in order to get a $ du $ in the numerator, I get
$ \int\frac{1}{1+e^x}\ dx = \int \frac{ e^x\,dx}{e^x(1+e^x)} = \int\frac{du}{(u-1)u}= \int\left[\frac{1}{u-1}-\frac{1}{u}\right]\ du $
via the method of Partial Fractions.
