Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$

Thanks to the power series representation of $1/(1+x^3)$ and the Dominated Convergence Theorem, the given integral is \begin{align*} -\int_0^1\frac{1-x^2}{(1+x^3)\log x}\,dx &=\int_0^1\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(1-x^2)x^{3k}}{\log x}\,dx\\ &=\int_0^1\sum_{k=0}^{\infty}\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\ &=\sum_{k=0}^{\infty}\int_0^1\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\&= \sum_{k=0}^{\infty}(\ln(6k+3)-\ln(6k+1)+\ln(6k+4)-\ln(6k+6))\\&= \ln\left(\prod_{k=0}^{\infty}\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right) =\ln\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right) \end{align*} where $\Gamma(x)$ is the Gamma function and we used the fact that $$\int_{0}^{1}\frac{x^n-1}{\log x}\,dx = \int_{0}^{+\infty}\frac{1-e^{-nt}}{t}\,e^{-t}dt = \log(n+1)$$ (apply Frullani's integral in the last step).

I was confused by the implicit use of the Frullani integral, so I think it bears mention: $$ \begin{align} \int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-mu}-e^{-nu}}{u}e^{-u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-(m+1)u}-e^{-(n+1)u}}{u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_{(m+1)\epsilon}^{(n+1)\epsilon}\frac{e^{-u}}{u}\,\mathrm{d}u\\[3pt] &=\log\left(\frac{n+1}{m+1}\right) \end{align} $$ $$ \begin{align} -\int_0^1\frac{1-x}{\left(1-x+x^2\right)\log(x)}\,\mathrm{d}x &=-\int_0^1\frac{1-x^2}{\left(1+x^3\right)\log(x)}\,\mathrm{d}x\\ &=-\sum_{k=0}^\infty(-1)^k\frac{x^{3k}-x^{3k+2}}{\log(x)}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\log\left(\frac{3k+3}{3k+1}\right)\\ &=\sum_{k=0}^\infty\log\left(\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right)\\ &=\log\left(\prod_{k=0}^\infty\frac{\left(k+\frac12\right)\left(k+\frac23\right)}{\left(k+\frac16\right)(k+1)}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=0}^{n-1}\frac{\color{#C00}{\left(k+\frac12\right)}\color{#090}{\left(k+\frac23\right)}}{\color{#00F}{\left(k+\frac16\right)}(k+1)}\right)\\ &=\log\left(\lim_{n\to\infty}\color{#C00}{\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)}}\color{#090}{\frac{\Gamma\left(n+\frac23\right)}{\Gamma\left(\frac23\right)}}\color{#00F}{\frac{\Gamma\left(\frac16\right)}{\Gamma\left(n+\frac16\right)}}\frac{\Gamma(1)}{\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right)+\log\left(\lim_{n\to\infty}\frac{\Gamma\left(n+\frac12\right)\Gamma\left(n+\frac23\right)}{\Gamma\left(n+\frac16\right)\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right) \end{align} $$ The last step is by Gautschi's Inequality.

I thought it might be instructive to present an approach that does not rely on Frullani's integral.

To that end, we first note that $\frac{x-1}{\log(x)}=\int_0^1 x^s\,ds$.

Therefore, we can write

$$\begin{align} \int_0^1 \frac{1+x}{1+x^3}\,\frac{x-1}{\log(x)}\,dx&=\int_0^1 \left(\int_0^1 \frac{x^s+x^{s+1}}{1+x^3} \right)\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n\int_0^1 (x^{s+3n}+x^{s+3n+1})\,dx\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n \left(\frac{1}{s+3n+1}+\frac{1}{s+3n+2}\right)\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+3}{3n+1}\right)\\\\ &=\sum_{n=0}^\infty \log\left(\frac{(6n+3)(6n+4)}{(6n+1)(6n+6)}\right)\tag 1\\\\ &=\log\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right)\tag2 \end{align}$$

where in going from $(1)$ to $(2)$ we relied on the analysis posted in Rob's solution herein.