integral (1/(e^x-1)) dx Using Partial Fractions
- Thread starter dagr8est
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skeeter
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I don't think this is really "partial fractions" in the sense that everyone knows them ...
\(\displaystyle \L \frac{1}{e^x - 1} = \frac{1 - e^x + e^x}{e^x - 1} = \frac{1 - e^x}{e^x - 1} + \frac{e^x}{e^x - 1} =\)
so ... \(\displaystyle \L \int \frac{e^x}{e^x - 1} - 1 dx = \ln(e^x - 1) - x + C\)
\(\displaystyle \L \frac{1}{e^x - 1} = \frac{1 - e^x + e^x}{e^x - 1} = \frac{1 - e^x}{e^x - 1} + \frac{e^x}{e^x - 1} =\)
so ... \(\displaystyle \L \int \frac{e^x}{e^x - 1} - 1 dx = \ln(e^x - 1) - x + C\)