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integral sign error |
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| Mar18-06, 05:16 PM | #1 |
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integral sign error
[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. |
| Mar18-06, 05:18 PM | #2 |
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For some reason it won't let me edit my post..... on the fifth line that integral should be
[tex]-\int\frac{du}{u} [/tex] |
| Mar18-06, 05:47 PM | #3 |
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Recognitions:
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sure, |a-b|=|b-a|
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| Mar18-06, 08:33 PM | #4 |
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integral sign error
ok thanks alot, stupid question
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