Mdtryhht

Inspired by the well known $$int_0^1fracln(1-x)ln xxmathrm dx=zeta(3)$$ and the integral given here (writing $zeta_r:=zeta(r)$ for easier reading)$$int_0^1fracln^3(1-x)ln xxmathrm dx=12zeta(5)-pi^2zeta(3)=6(2zeta_5-zeta_3zeta_2),$$ I have looked at the integrals
$$I_n,m=:int_0^1fracln^n(1-x)ln^m xxmathrm dx$$ for $m,ninmathbb N$.

From the power series of $ln(1-x)$ and recursion, it is not hard to see that $I_m,n$ can be written as a rational combination of products of integer zeta values, with the arguments ($ge 2$) for each product summing up to $m+n+1$, e.g. $$I_n,1=begincases -n!Bigl(dfracn-14zeta_n+2-dfrac12sumlimits_j=1^k-1zeta_2j+1zeta_n+1-2j Bigr) &text for n=2k\
-n!Bigl(dfracn+12zeta_n+2- sumlimits_j=1^k-1zeta_2j+1zeta_n+1-2jBigr) &text for n=2k-1 endcases.$$

So far, so good. Now the intriguing (numerical) discovery is that there are several families of pairs of those integrals that have rational ratios. For instance, $$fracI_n+2,n-2I_n-1,n+1=fracn+2n-1quad textorquadfracI_n+1,n-1I_n,n=fracn+1nquad textorquadfracI_n,2I_3,n-1=fracn3,$$ e.g. $$I_5,2=colorred10colorblue(61zeta_8-72zeta_5zeta_3+12zeta_3^2zeta_2) $$ and
$$I_3,4=colorred6colorblue(61zeta_8-72zeta_5zeta_3+12zeta_3^2zeta_2). $$ Note that each such pair has the same proportion as the respective first arguments.

I don't see how that could be possibly proven by recursion, so there must be some deeper connection between those integrals.

How to explain that?

For $ngeq 1$ and $mgeq 0$, an application of integration by parts ($u=log^n(1-x)$, $dv=log^m(x),dx/x$) followed by the substitution $xmapsto 1-x$ shows that
$$
fracI_n,mI_m+1,n-1=fracnm+1.
$$
All of your examples are special cases of this identity.