While playing around with the first values of the integral
$$ I_n:=-\int_0^1\left(x(1-x)\right)^n\frac{d^n}{d^nx}\left(\log x \cdot\log (1-x)\right){\rm d}x, \quad \quad n=1,2,3,\cdots, $$
I got $$ \small{\begin{align} I_1&=0,&I_2&=\frac19,&I_3&=0,&I_4&=\frac3{25},\\ I_5&=0,&I_6&=\frac{40}{49},&I_7&=0,&I_8&=\frac{140}{9},\\ I_9&=0,&I_{10}&=\frac{72576}{121},&I_{11}&=0,&I_{12}&=\frac{6652800}{169},\\ I_{13}&=0,&I_{14}&=\color{#99004d}{3953664},&I_{15}&=0,&I_{16}&=\frac{163459296000}{289},\\ I_{17}&=0,&I_{18}&=\frac{39520825344000}{361},&I_{19}&=0,&I_{20}&=\color{#99004d}{27583922995200},\\ I_{21}&=0,&I_{22}&=\frac{4644631106519040000}{529},&I_{23}&=0,&I_{24}&=\color{#99004d}{3446935565184663552},\\ I_{25}&=0,&I_{26}&=\color{#99004d}{1636721540923392000000},&I_{27}&=0,&I_{28}&=\frac{777776389315596582912000000}{841},\\ I_{29}&=0,&I_{30}&=\cdots. \end{align}} $$
By splitting up the initial integral into $\displaystyle \int_0^{1/2}$, $\displaystyle \int_{1/2}^1$ and by using the symmetry of the integrand, I've indeed proved that $I_{2n+1}=0, \, n=0,1,2,3,\cdots.$
Now observing the first values above, my question is:
Does the integral $I_{2n}$ take on infinitely integer values?