Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$ I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate  
$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$ 
Wolfram Alpha online calculator provide me a closed-form with code
int -(1-x)/((1-x+x^2)log(x)) dx, from x=0 to x=1

Question. Please provide me hints to know how evaluate previous this definite integral as
  $$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx=\log \left(\frac{\Gamma(1/6)}{\Gamma(2/3)}\right)-\frac{\log \pi}{2}$$
  as said Wolfram Alpha. Many thanks.

 A: 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.
A: 
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.
A: 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).
A: Let
$$I=\int_0^1\frac{1+x}{1+x^3}\frac{x-1}{\ln(x)}dx$$
and
$$I(a)=\int_0^1\frac{1+x}{1+x^3}\frac{x^a-1}{\ln(x)}dx$$
and notice that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^a+x^{a+1}}{1+x^3}dx\overset{x=t^{1/3}}{=}\frac13\int_0^1\frac{t^{\frac{a-2}{3}}+t^{\frac{a-1}{3}}}{1+t}dt$$
By using
$$\int_0^1\frac{x^n}{1+x}dx=\frac12\psi\left(\frac{n+2}{2}\right)-\frac12\psi\left(\frac{n+1}{2}\right),$$
we have
$$I'(a)=\frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right].$$
Integrate both sides from $a=0$ to $a=1$
$$\int_0^1 I'(a)da=I(a)|_0^1=I(1)-I(0)=I-0$$
$$=\int_0^1 \frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right]da$$
$$=\left[\ln\Gamma\left(\frac{a+4}{6}\right)-\ln\Gamma\left(\frac{a+1}{6}\right)+\ln\Gamma\left(\frac{a+5}{6}\right)-\ln\Gamma\left(\frac{a+2}{6}\right)\right]_0^1$$
$$=\left[\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)+\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{1}{2}\right)\right]$$
$$-\left[\ln\Gamma\left(\frac{2}{3}\right)-\ln\Gamma\left(\frac{1}{6}\right)+\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)\right]$$
$$=-\ln\Gamma\left(\frac{1}{2}\right)-\ln\Gamma\left(\frac{2}{3}\right)+\ln\Gamma\left(\frac{1}{6}\right)$$
$$=\ln\left(\frac{\Gamma\left(\frac16\right)}{\sqrt{\pi}\, \Gamma\left(\frac23\right)}\right).$$
