Show that $\int_{a}^{b}\frac{\ln x ~ dx}{(x+a)(x+b)}=\frac{\ln ab}{2(b-a)} \ln \left(\frac{(a+b)^2}{4ab} \right), a, b>0.$ This integral is tabulated in Gradshetyn and Ryzhik.
The  Mathematica gives nice results only for particular values of $a$ and $b$. Can one prove this result?
 A: Following MartinR's suggestion we show a slight generalization. 
Let $r>0$ and $0<a<b$. By using the substitution $t=\frac{ab}{x}$, it follows that
$$\begin{align}
I_r&:=\int_{a/r}^{rb}\frac{\ln(x) }{(x+a)(x+b)}\,dx\\
&=\int_{rb}^{a/r}\frac{\ln(ab/t) }{(\frac{ab}{t}+a)(\frac{ab}{t}+b)}\cdot -\frac{ab}{t^2}dt\\
&=
\int_{a/r}^{rb}\frac{\ln (ab/t) }{(b+t)(a+t)}\,dt\\&=\ln (ab)\int_{a/r}^{rb}\frac{dt}{(b+t)(a+t)}\,dt-I.\end{align}$$
Hence
$$\begin{align}
I_r&=\frac{\ln (ab)}{2}\int_{a/r}^{rb}\frac{dt}{(b+t)(a+t)}\,dt\\
&=\frac{\ln (ab)}{2(b-a)}\int_{a/r}^{rb}\left(\frac{1}{a+t}-\frac{1}{b+t}\right)\,dt
\\&=\frac{\ln (ab)}{2(b-a)}\left[\ln\left(\frac{a+t}{b+t}\right)\right]_{a/r}^{rb}\\
&=
\frac{\ln (ab)}{2(b-a)}\ln\left(\frac{(a+rb)^2}{(1+r)^2ab}\right).
\end{align}$$
For $r=1$ we got the result. For $r\to +\infty$ we obtain
$$\int_0^{+\infty} \frac{ \ln(x)}{(x+a)(x+b)} dx=\frac{\ln (ab)\ln(b/a)}{2(b-a)}=\frac{\ln^2 (b)-\ln^2(a)}{2(b-a)}.$$
See Integral involving logarithm: $\int_0^\infty \frac{ \ln x}{(x+a)(x+b)} dx$.
P.S. Case $b=-1$ and $a>0$
$$\begin{align}
\int_0^1 \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx 
&= \frac{1}{1+a}\int_0^1 \frac{\ln x}{1-x}dx - \frac{1}{1+a}\int_0^{1}\frac{\ln (x)}{x+a}dx \\
&= \frac{\pi^2}{6(1+a)}-\frac{\text{Li}_2(-1/a)}{1+a}
\end{align}$$
Moreover, by letting $t=1/x$
$$\begin{align}\int_1^{+\infty} \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx
&=\frac{1}{a}\int_0^{1} \frac{\ln{t}}{(1-t)(1/a+t)}dx\\
&=\frac{1}{a}\left(\frac{\pi^2}{6(1+1/a)}-\frac{\text{Li}_2(-a)}{1+1/a}\right)
\end{align}$$
Hence, after summing the two parts we get
$$\int_0^{+\infty} \frac{\ln{x}}{\left(1-x\right)\left(x+a\right)}dx=\frac{\pi^2}{3(1+a)}-\frac{\text{Li}_2(-1/a)+\text{Li}_2(-a)}{1+a}=\frac{\pi^2+\ln^2(a)}{2(a+1)}$$
where we used the identity $\text{Li}_2(-1/a)+\text{Li}_2(-a)=-\frac{\pi^2}{6}-\frac{\ln^2(a)}{2}$.
