Let $a$ and $b$ be greater than $0$ and let
$$I=\int_0^{\infty}\frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$
I know I can first manipulate by adding and subtracting $1$ in the numerator. Thus
$$\int_0^{\infty}\frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^{\infty}\frac{1+e^{ax}-1-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^{\infty}\frac{(1+e^{ax})-(1+e^{bx})}{(1+e^{ax})(1+e^{bx})}dx$$
and then we can split it into two integrals
$$=\int_0^{\infty}\frac{1}{1+e^{bx}}dx-\int_0^{\infty}\frac{1}{1+e^{ax}}$$
Let $u=bx$, $v=ax$. Then $du=bdx$ and $dv=adx$. This yields
$$=\frac{1}{b}\int_0^{\infty}\frac{du}{1+e^{u}}-\frac{1}{a}\int_0^{\infty}\frac{dv}{1+e^{v}}$$
Now, let $s=1+e^u$, $t=1+e^v$. Then $ds=e^udu, dt=e^vdv$. But since $e^u=s-1$ and $e^v=t-1$, we get as integrals
$$=\frac{1}{b}\int_2^{\infty}\frac{ds}{s(s-1)}-\frac{1}{a}\int_2^{\infty}\frac{dt}{t(t-1)}$$
This can be then split using PFD and the result is
$$=\frac{1}{b}\int_2^{\infty}\frac{ds}{s-1}-\frac{1}{b}\int_2^{\infty}\frac{ds}{s}-\frac{1}{a}\int_2^{\infty}\frac{dt}{t-1}+\frac{1}{a}\int_2^{\infty}\frac{dt}{t}$$
Thus
$$I=\left[\frac{1}{b}\ln(s)-\frac{1}{b}\ln(s-1)\right]_2^{\infty}-\left[\frac{1}{a}\ln(t)-\frac{1}{a}\ln(t-1)\right]_2^{\infty}$$ $$=\lim_{n\rightarrow\infty}\frac{1}{b}\left[\ln\left(\frac{s}{s-1}\right)\right]_2^n-\lim_{n\rightarrow\infty}\frac{1}{a}\left[\ln\left(\frac{t}{t-1}\right)\right]_2^n$$ $$=\frac{1}{b}[0-\ln(2)]-\frac{1}{a}[0-\ln(2)]$$ $$=\left(\frac{1}{a}-\frac{1}{b}\right)\ln(2)$$ $$=\left(\frac{a-b}{ab}\right)\ln(2)$$
So I feel pretty confident this is correct. My question is this. The question required multiple substitutions in order to solve this. Is there any better methods for calculating $I$ that are quicker or cleaner? Is there a more obvious approach as opposed to the multiple subsitituion approach?
