- #1
- 2,704
- 19
[tex] \int \frac{1+e^x}{1-e^x}dx = [/tex]
[tex] \int \frac{dx}{1-e^x} + \int \frac{e^x}{1-e^x}dx [/tex]
The second integral can be done by the substitution
[tex] u = 1-e^x [/tex]
[tex] du = -e^x dx[/tex]
So the second integral becomes:
[tex]\int \frac{du}{u} = \ln|u|+C[/tex]
In the first integral, you can use the substituion:
[tex]w = e^x[/tex]
[tex]dw = e^x dx[/tex]
So the first integral becomes:
[tex] \int \frac{dw}{w(1-w)} [/tex]
This can be done by parts and you get:
[tex] \int \frac{dw}{w} + \int \frac{dw}{1-w} [/tex]
These are also both natural logs so you end up with:
[tex] \ln|e^x| - 2\ln|1-e^x| + C [/tex]
[tex] x - 2\ln|1-e^x| +C [/tex]
I had http://integrals.wolfram.com/index.jsp compute this for me and it got:
[tex] x - 2\ln(e^x - 1) +C [/tex]
I think this may have something to do with the absolute value, but I always make stupid mistakes with signs so I thought I'd check...Thanks for the time I probably just wasted.
[tex] \int \frac{dx}{1-e^x} + \int \frac{e^x}{1-e^x}dx [/tex]
The second integral can be done by the substitution
[tex] u = 1-e^x [/tex]
[tex] du = -e^x dx[/tex]
So the second integral becomes:
[tex]\int \frac{du}{u} = \ln|u|+C[/tex]
In the first integral, you can use the substituion:
[tex]w = e^x[/tex]
[tex]dw = e^x dx[/tex]
So the first integral becomes:
[tex] \int \frac{dw}{w(1-w)} [/tex]
This can be done by parts and you get:
[tex] \int \frac{dw}{w} + \int \frac{dw}{1-w} [/tex]
These are also both natural logs so you end up with:
[tex] \ln|e^x| - 2\ln|1-e^x| + C [/tex]
[tex] x - 2\ln|1-e^x| +C [/tex]
I had http://integrals.wolfram.com/index.jsp compute this for me and it got:
[tex] x - 2\ln(e^x - 1) +C [/tex]
I think this may have something to do with the absolute value, but I always make stupid mistakes with signs so I thought I'd check...Thanks for the time I probably just wasted.