Integrate $\int_0^1 \ln(x)\ln(b-x)\,\mathrm{d}x$, for $b>1$? Let $b>1$. What's the analytical expression for the following integral?
$$\int_0^1 \ln(x)\ln(b-x)\,\mathrm{d}x$$
Mathematica returns the following answer:
$$2-\frac{\pi^{2}}{3}b+\left(b-1\right)\ln\left(b-1\right)-b\ln b+\mathrm{i}b\pi\ln b+\frac{1}{2}b\ln^{2}b+b\mathrm{Li}_{2}\left(b\right)$$
which contains the imaginary term $\mathrm{i}b\pi\ln b$. But the actual answer is real, so this term should cancel somehow with the dilogarithm function. But I don't know how to do this.

Define
$$ I(a)=\int_0^1x^a\ln(b-x)dx. $$
It is easy to see that
$$ \int_0^1\ln x\ln(b-x)dx=\lim_{a\to 0^+}I'(a). $$
Now
\begin{eqnarray*}
I(a)&=&\frac{1}{a+1}\int_0^1\ln(b-x)d(x^{a+1})=\frac{1}{a+1}\left(x^{a+1}\ln(b-x)|_0^1+\int_0^1\frac{x^{a+1}}{b-x}dx\right)\\
&=&\frac{1}{a+1}\left(\ln(b-1)+\frac{1}{b}\sum_{n=0}^\infty\int_0^1\frac{1}{b^n}x^{a+1+n}dx\right)\\
&=&\frac{1}{a+1}\left(\ln(b-1)+\sum_{n=0}^\infty\frac{1}{b^{n+1}(a+2+n)}\right)\\
\end{eqnarray*}
and hence
\begin{eqnarray*}
I'(0)
&=&-\ln(b-1)-\sum_{n=0}^\infty\frac{1}{b^{n+1}(2+n)}-\sum_{n=0}^\infty\frac{1}{b^{n+1}(2+n)^2}\\
&=&-\ln(b-1)+2+b\ln\frac{b-1}{b}-b\text{Li}_2(1/b).
\end{eqnarray*}
