We first make the observation that
$\displaystyle \displaystyle \frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\int_a^b\frac{dy}{(x+y)^2} \text{ }dy$
And thus, if $\displaystyle I(a,b)$ is the given integral we have that
$\displaystyle \displaystyle I(a,b)=\frac{1}{b-a}\int_0^\infty \int_a^b \frac{\log(x)}{(x+y)^2}\text{ }dy\,dx$
Now, since
$\displaystyle \displaystyle \int_{\mathbb{R}\times[a,b]}\left|\frac{\log(x)}{(x+y)^2}\right|\text{ }d(x,y)<\infty$
(which is easily verified via estimation) we may apply Fubini's theorem to conclude that
$\displaystyle \displaystyle I(a,b)=\frac{1}{b-a}\int_a^b\int_0^\infty \frac{\log(x)}{(x+y)^2}\text{ }dx\,dy$
But, it is easy to see that
$\displaystyle \displaystyle \int\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{x\log(x)}{y(y+x)}-\frac{\log(y+x)}{y}$
So that
$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{\log(y)}{y}$
and so
$\displaystyle \begin{aligned}I(a,b) &=\frac{1}{b-a}\int_a^b\int_0^\infty\frac{\log(x)}{(x+y)^2} \text{ }dx\\ &=\frac{1}{b-a}\int_a^b\frac{\log(y)}{y}\;dy\\ &=\frac{\log^2(b)-\log^2(a)}{2(b-a)}\end{aligned}$