| Line 7: | Line 7: | ||
Alright, I factored out a <math> e^x </math>. I set 'u' equal to <math> 1+\frac{1}{e^x} </math> and 'du' to <math> -1\frac{1}{e^x} dx </math>. The problem looks like this: <math> \int \frac{dx}{e^x(1+\frac{1}{e^x})} </math>. Substitute with u and du. I got this answer: <math> -Ln|1+\frac{1}{e^x}| + C </math> | Alright, I factored out a <math> e^x </math>. I set 'u' equal to <math> 1+\frac{1}{e^x} </math> and 'du' to <math> -1\frac{1}{e^x} dx </math>. The problem looks like this: <math> \int \frac{dx}{e^x(1+\frac{1}{e^x})} </math>. Substitute with u and du. I got this answer: <math> -Ln|1+\frac{1}{e^x}| + C </math> | ||
| + | |||
| + | Here's another way. Let <math>u=1+e^x</math>. Then <math>du=e^x dx</math> and <math>e^x=u-1</math>. If I multiply | ||
| + | and divide by <math>e^x</math> in order to get a <math>du</math> in the numerator, I get | ||
| + | |||
| + | <math>\int\frac{1}{1+e^x}\ dx = \int \frac{ e^x\,dx}{e^x(1+e^x) = \int\frac{du}{(1-u)u}</math> | ||
| + | |||
| + | and this last integral can be computed via integration by parts. | ||
Revision as of 05:40, 30 September 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 $
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}{(1-u)u} $
and this last integral can be computed via integration by parts.
