How can I solve this Integral?

[Alexandra]

Homework Statement

Hi, I'm trying to solve the next integral

Relevant Equations

\begin{equation}
\int ^{\infty} _{0} \dfrac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})} \ dx
\end{equation}
a > 0 ; b > 0

I split this to get
\begin{equation}
\int ^{\infty} _{0} \dfrac{e^{ax}}{(1+e^{ax})(1+e^{bx})} \ dx - \int ^{\infty} _{0} \dfrac{e^{bx}}{(1+e^{ax})(1+e^{bx})} \ dx
\end{equation}
Then I tried to solve the first term (both term are similars). The problem is that I made a substitution (many ones, but this has, for me, more meaning), but it didn't work: if u=exp(a*x), then
\begin{equation}
\dfrac{1}{a} \int ^{\infty} _{0} \dfrac{1}{(1+u)(1+u^{b/a})} \ dx
\end{equation}
I can't do partial fraction decomposition to this (b/a couldn't be a natural number).
I really don't know how to solve this. I put this into Wolfram Mathematica, but it can't solve it. I forgot mention that this is a multiple choice exercise. The posible answers are: 0 , 1 , b-a , (a-b) log 2 , [(a-b)/ab] log 2.
I suposse that 0 can't be the answer, because the solution of each term on (3) should depende on a and b by the same way (I know, it isn't a good enough reason).
1 neither could be a solution, because the terms in (3) are "simetrics". If the solution of one of then doesn't depend of a or b, then both terms are equal and the solution would be zero.
Then the solution would be some of the other 3 options, but I don't know which arguments are valid to say which one.
I hope someone can help me. Sorry for my bad english.
Thanks!

 

[Alexandra]

Sorry, I had two mistakes. When I reference the equation (3), I meant equation (2). And the lower index on the integral (3) is 1, not 0.

 

[epenguin]

Isn't this something relatively trivial?:

$$\dfrac {A-B}{\left( 1+A\right) \left( 1+B\right) }=\dfrac {1}{\left( 1+B\right) }-\dfrac {1}{\left( 1+A\right) }$$

 

[Alexandra]

Now I feel very stupid...Thanks epenguin!

 

[vela]

Another trick you could use is tweaking the numerator as follows:
$$\int^\infty_0 \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})}\,dx = \int^\infty_0 \frac{(1+e^{ax}) - (1+e^{bx})}{(1+e^{ax})(1+e^{bx})}\,dx$$

 

[Alexandra]

Another trick you could use is tweaking the numerator as follows:
$$\int^\infty_0 \frac{e^{ax} - e^{bx}}{(1+e^{ax})(1+e^{bx})}\,dx = \int^\infty_0 \frac{(1+e^{ax}) - (1+e^{bx})}{(1+e^{ax})(1+e^{bx})}\,dx$$

That 's a very good trick, thanks vela!