<s>[INST] Prove $I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$ How can we show that:
$$I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$$
The logarithm in the denominator is the main trouble for me here.
I have obtained this integral from the following infinite product:
$$P=e^I=\prod_{k=1}^\infty \frac{(3k-2)^2(6k-1)}{(3k-1)^2(6k-5)}=\sqrt[3]{2}$$
Which can probably be proved by Gamma functions, though I don't have the proof right now.
Is there a way to evaluate the integral without going back to the product in question? Preferably by some semi-elementary means.
A more simple, but related integral is:
$$\int_0^1 \frac{(x-1) dx}{\ln x}=\ln 2$$
I'm sure it's here somewhere, so maybe the same methods of solution would work for the more complicated case.

Update:
I found a way to transform the integral which makes it look even more complicated. If:
$$f(s)=\frac{2 }{s+1}\left(\frac{\, _2F_1\left(1,\frac{s+1}{2};\frac{s+3}{2};-\frac{1}{2} \left(i \sqrt{3}+1\right)\right)}{3+i \sqrt{3}}+\frac{\, _2F_1\left(1,\frac{s+1}{2};\frac{s+3}{2};-\frac{1}{2} \left(1-i \sqrt{3}\right)\right)}{3-i \sqrt{3}}\right)$$
Then:
$$I=\int_0^1 (f(s)-f(s+1)) ds$$
 [/INST] By using the integral representation
\begin{equation}
\frac{x-1}{\ln x}=\int_0^1x^s\,ds
\end{equation} 
we can express
\begin{align}
I&=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}\\
&=\int_0^1 \,ds\int_0^1 \frac{x^s\left( 1-x \right)}{1+x^2+x^4}\,dx
\end{align}
or, with the notation $J(s)=\int_0^1 \frac{x^s}{1+x^2+x^4}\,dx$,
\begin{equation}
I=\int_0^1 \left( J(s)-J(s+1) \right)\,ds
\end{equation} 
By changing $x=y^2$ and denoting $j=\exp(2i\pi/3)$,
\begin{align}
J(s)&=\frac{1}{2}\int_0^1\frac{y^{\frac{s-1}{2}}}{1+y+y^2}\,dy\\
&=\frac{1}{2j(1-j)}\int_0^1y^{\frac{s-1}{2}}\left[\frac{j}{1-jy}-\frac{j^2}{1-j^2y}\right]\,dy\\
&=\frac{1}{\sqrt{3}}\Im\int_0^1y^{\frac{s-1}{2}}\frac{j}{1-jy}\,dy
\end{align}
Expansion of $(1-jy)^{-1}$ gives
\begin{align}
J(s)&=\frac{1}{\sqrt{3}}\Im\int_0^1 \sum_{n\ge0}j^{n+1}y^{\frac{s-1}{2}+n}\,dy\\
&=\frac{1}{\sqrt{3}}\Im\sum_{n\ge0}\frac{j^{n+1}}{\frac{s+1}{2}+n}
\end{align}
As $j^3=1$, $\Im j=\sqrt{3}/2$ and $\Im j^2=-\sqrt{3}/2$
\begin{align}
J(s)&=\frac{2}{\sqrt{3}}\Im\sum_{p\ge0}\left[\frac{j}{s+1+6p}+\frac{j^2}{s+3+6p}\right]\\
&=\sum_{p\ge0}\left[\frac{1}{s+1+6p}-\frac{1}{s+3+6p}\right]\\
\end{align} 
We can use the representation
\begin{equation}
\psi(a)-\psi(b)=\sum_{k=0}^\infty\left[\frac{1}{k+b}-\frac{1}{k+a}\right]
\end{equation} 
where $\Psi(.)$ is the digamma function, to express
\begin{equation}
J(s)=\frac{1}{6}\left[\Psi\left( \frac{s+3}{6} \right)-\Psi\left( \frac{s+1}{6} \right)\right]
\end{equation} 
We have thus to evaluate
\begin{align}
I&=\frac{1}{6}\int_0^1\left[\Psi\left( \frac{s+3}{6} \right)-\Psi\left( \frac{s+1}{6} \right)-\Psi\left( \frac{s+4}{6} \right)+\Psi\left( \frac{s+2}{6} \right)\right]\,ds\\
&=\ln\left[\frac{\Gamma(2/3)}{\Gamma(1/2)}\right]-\ln\left[\frac{\Gamma(1/3)}{\Gamma(1/6)}\right]-\ln\left[\frac{\Gamma(5/6)}{\Gamma(2/3)}\right]+\ln\left[\frac{\Gamma(1/2)}{\Gamma(1/3)}\right]\\
&=\ln\left[\frac{\Gamma^2(2/3)\Gamma(1/6)}{\Gamma^2(1/3)\Gamma(5/6)}\right]
\end{align} 
But, using the duplication formula for the Gamma function
\begin{equation}
\frac{\Gamma^2(2/3)}{\Gamma^2(1/3)}=\frac{2^{-2/3}}{\pi}\Gamma^2(5/6)
\end{equation} 
and the reflection formula
\begin{equation}
\Gamma(5/6)\Gamma(1/6)=\frac{\pi}{\sin\pi/6}
\end{equation} 
we deduce
\begin{equation}
I=\frac{\ln2}{3}
\end{equation} 
</s>