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A variation on: Another Integral involving $e^{ax} +1$ and $e^{bx} + 1$

Evaluate the integral $$I(a,b)=\int_{0}^{1}\frac{(e^{ax})(e^{bx})}{\left(e^{ax}+1\right)\left(e^{bx}+1\right)}dx$$ for $a>b>0$.

Attempt

I suppose, like before, I have to simplify the integrand to seperate the $a$ and $b$ into different integrals. So far I have managed to do this:

$$1 - \frac{1}{\left(e^{bx}+1\right)} - \frac{(e^{bx})}{\left(e^{ax}+1\right)\left(e^{bx}+1\right)}$$

But I am stuck. Not sure if the question is unsolvable this way or if I can't see the trick.

Additional Info

This is not an "official" question from a textbook, course or quiz. There might be no nice solutions.

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    So you are sure there is a "good" answer here. For example mathematica 8 can't find explicit formula here.2012-11-03
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    @Norbert - Nope, I am not sure if there is a "good" answer. Perhaps there is no nice solution. Thanks for the input.2012-11-03
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    Then the mention *Just like before, this is a "putnam practice" type of question that is meant to be solved in less than 5 minutes using "simple" mathematics* is quite misleading.2012-11-03
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    @did - Sorry about that! I'll edit it. (its from a set of old second hand notes containing "putnam" type practice questions, parts of which are quite illegible)2012-11-03

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Hint: $$I(a,b)=\int_{0}^{1}1-\frac{1}{\left(e^{bx}+1\right)}-\frac{1}{\left(e^{ax}+1\right)}+\frac{1}{\left(e^{ax}+1\right)\left(e^{bx}+1\right)}dx.$$

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    What about the fourth term?2012-11-03
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    I reached this stage before but is unsure about how to handle the fourth term too.2012-11-03
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    Based on the comments in the question, this appears to be the best possible result. Thanks.2012-11-05