Inspired by the well known $$\int\_0^1\frac{\ln(1-x)\ln x}x\mathrm dx=\zeta(3)$$ and the integral given [here](https://math.stackexchange.com/questions/908108/how-to-find-large-int-01-frac-ln31x-ln-xx-mathrm-dx) (writing $\zeta\_r:=\zeta(r)$ for easier reading)$$\int\_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx=12\zeta(5)-\pi^2\zeta(3)=6(2\zeta\_5-\zeta\_3\zeta\_2),$$ I have looked at the integrals
$$I\_{n,m}=:\int\_0^1\frac{\ln^n(1-x)\ln^m x}x\mathrm dx$$ for $m,n\in\mathbb N$.


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. $$I\_{n,1}=\begin{cases} -n!\Bigl(\dfrac{n-1}4\zeta\_{n+2}-\dfrac12\sum\limits\_{j=1}^{k-1}\zeta\_{2j+1}\zeta\_{n+1-2j} \Bigr)\ \ &\text{ for }n=2k, \\
-n!\Bigl(\dfrac{n+1}2\zeta\_{n+2}- \sum\limits\_{j=1}^{k-1} \zeta\_{2j+1} \zeta\_{n+1-2j}\Bigr)\ \ &\text{ for }n=2k-1. \end{cases}$$


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, $$\frac{I\_{n+2,n-2}}{I\_{n-1,n+1}}=\frac{n+2}{n-1}\quad \text{or}\quad\frac{I\_{n+1,n-1}}{I\_{n,n}}=\frac{n+1}{n}\quad \text{or}\quad\frac{I\_{n,2}}{I\_{3,n-1}}=\frac{n}{3},$$ e.g. $$I\_{5,2}=\color{red}{10}\color{blue}{(61\zeta\_8-72\zeta\_5\zeta\_3+12\zeta\_3^2\zeta\_2)} $$ and
$$I\_{3,4}=\color{red}6\color{blue}{(61\zeta\_8-72\zeta\_5\zeta\_3+12\zeta\_3^2\zeta\_2)}. $$ 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?
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> 
> 



Why these surprising proportionalities of integrals involving odd zeta values?
For $n\geq 1$ and $m\geq 0$, an application of integration by parts ($u=\log^n(1-x)$, $dv=\log^m(x)\,dx/x$) followed by the substitution $x\mapsto 1-x$ shows that
$$
\frac{I\_{n,m}}{I\_{m+1,n-1}}=\frac{n}{m+1}.
$$
All of your examples are special cases of this identity.


