Integral $\int_0^1(x(1-x))^n\frac{d^n}{d^n x}(\log x \cdot\log (1-x))dx$

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?

1) We have first with $\displaystyle D=\frac{d}{dx}$:

$$J_n(x)=D^{n}(\log(x)\log(1-x))=\sum_{k=0}^{n}{n\choose k}D^{n-k}(\log(x))D^{k}(\log(1-x))$$ hence $$J_n(x)=D^{n}(\log(x))\log(1-x)+D^{n}(\log(1-x))\log (x)+\sum_{k=1}^{n-1}{n\choose k}D^{n-k}(\log(x))D^{k}(\log(1-x))$$

Now for $m\geq 1$ $$D^{m}(\log(x))=D^{m-1}(1/x)=(-1)^{m-1}\frac{(m-1)!}{x^m}$$ and $$D^{m}(\log(1-x))=-D^{m-1}(1/(1-x))=-\frac{(m-1)!}{(1-x)^m}$$

This gives $$J_n(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}\log(1-x)-\frac{(n-1)!}{(1-x)^n}\log x+\sum_{k=1}^{n-1}{n\choose k}(-1)^{n-k}\frac{(k-1)!(n-k-1)!}{x^{n-k}(1-x)^{k}}$$

2) We multiply by $x^n(1-x)^n$, and we note that $$\int_0^1 x^n \log (x)dx=\int_0^1(1-x)^n\log(1-x)dx=-\frac{1}{(n+1)^2}$$ and $$\int_0^1 x^k(1-x)^{n-k}dx=B(k+1,n-k+1)=\frac{k!(n-k)!}{(n+1)!}$$ we find that $$-I_n=(1-(-1)^{n-1})\frac{(n-1)!}{(n+1)^2}+\frac{1}{n+1}\sum_{k=1}^{n-1}(-1)^{n-k}(k-1)!(n-k-1)! $$

3) If $n=2m+1$, we have $$(2m+2)I_{2m+1}=\sum_{k=1}^{2m}(-1)^k (k-1)!(2m-k)!=B_m$$

The change of parameter $k'=2m+1-k$ give $B_m=-B_m$, hence $B_m=0$.

4) If $n=2m$, we have

$$-I_{2m}=2\frac{(2m-1)!}{(2m+1)^2}+\frac{1}{2m+1}\sum_{k=1}^{2m-1}(-1)^k (k-1)!(2m-1-k)! $$

Note that $(m-1)!$ divide $(k-1)!(2m-1-k)!$ for all $k$, $1\leq k\leq 2m-1$. So to have $I_{2m}\in \mathbb{Z}$, it suffice to have that $(2m+1)^2$ divide $(2m-1)!$ and $2m+1$ divide $(m-1)!$;

This is true for example if for $m\geq 2$, we have $2m+1=p_1p_2$ with $p_1$, $p_2$ prime with $p_1<p_2$, as this imply that $3\leq p_1<p_2<m$, hence $p_1^2$ and $p_2^2$ divide $(2m-1)!$, and $p_1$, $p_2$ divide $(m-1)!$

So for all these values of $m$, we have $I_{2m}\in \mathbb{Z}$.