# 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 $a>b>0$.

Just like the related question, this is a "putnam practice" type of question that is meant to be solved in less than 5 minutes using "simple" mathematics.

I am not sure if the question is actually solvable. There might be crucial details missing (this was found scribbled on old notes).

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Hint: $$\frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}=\frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}$$

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This is embarrassingly straight forward. :S Thanks for the reminder! –  Legendre Nov 1 '12 at 13:33

Hint: $$\frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}=\frac{e^{ax}}{1+e^{ax}}-\frac{e^{bx}}{1+e^{bx}}$$

Take an educated guess.

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Hint: $$I(a,b)=\int_0^1\frac{(e^{ax}+1)-(e^{bx}+1)}{(e^{ax}+1)(e^{bx}+1)}dx=\int_0^1\frac{dx}{e^{bx}+1}-\int_0^1\frac{dx}{e^{ax}+1}$$ Now $$\int_0^1\frac{dx}{e^{bx}+1}=\int_0^1\frac{1}{b}\frac{d(e^{bx})}{(e^{bx})^2+e^{bx}}=\frac{1}{b}\int_1^{e^b}\frac{dz}{z^2+z}=\frac{1}{b}\int_1^{e^b}\left(\frac{1}{z}-\frac {1}{z+1}\right)dz$$ and this is a well known integral and can be solved

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Thanks for the help! +1 –  Legendre Nov 1 '12 at 13:33