[tex] \int \frac{1+e^x}{1-e^x}dx = [/tex](adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Homework Help: Integral sign error

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