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\begin{align}
&-\int_{0}^{1}{1 - x \over \pars{1 - x + x^{2}}\ln\pars{x}}\,\dd x =
\int_{0}^{1}\
\overbrace{{1 + x \over 1 + x^{3}}}
^{\ds{1 \over 1 - x + x^{2}}}\,\,\,
\overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{\int_{0}^{1}x^{t}\,\dd t}}\
\,\dd x =
\int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} \over 1 + x^{3}}\,\dd x\,\dd t
\\[5mm] & =
\int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} - x^{t + 3} - x^{t + 4} \over
1 - x^{6}}\,\dd x\,\dd t
\\[5mm] & \stackrel{x^{6}\ \mapsto\ x}{=}\,\,\,
{1 \over 6}\int_{0}^{1}\int_{0}^{1}
{x^{t/6 - 5/6} + x^{t/6 - 2/3} - x^{t/6 - 1/3} - x^{t/6 - 1/6} \over
1 - x}\,\dd x\,\dd t
\\[5mm] & =
{1 \over 6}\int_{0}^{1}\!\!\pars{%
-\int_{0}^{1}\!\!{1 - x^{t/6 - 5/6} \over 1 - x}\,\dd x -
\int_{0}^{1}\!\!{1 - x^{t/6 - 2/3} \over 1 - x}\,\dd x +
\int_{0}^{1}\!\!{1 - x^{t/6 - 1/3} \over 1 - x}\,\dd x +
\int_{0}^{1}\!\!{1 - x^{t/6 - 1/6} \over 1 - x}\,\dd x}
\\[5mm] & =
{1 \over 6}\ \overbrace{\int_{0}^{1}\bracks{%
-\Psi\pars{{t \over 6} + {1 \over 6}} -
\Psi\pars{{t \over 6} + {1 \over 3}} +
\Psi\pars{{t \over 6} + {2 \over 3}} +
\Psi\pars{{t \over 6} + {5 \over 6}}}\dd t}
^{\ds{\Psi:\ Digamma\ Function}}
\\[5mm] & =
\left.\ln\pars{\Gamma\pars{t/6 + 2/3}\Gamma\pars{t/6 + 5/6} \over \Gamma\pars{t/6 + 1/6}\Gamma\pars{t/6 + 1/3}}\right\vert_{\ 0}^{\ 1}\qquad
\pars{~\Gamma:\ Gamma\ Function.\ \Psi\pars{z}
\stackrel{\mrm{def.}}{=} \totald{\ln\pars{\Gamma\pars{z}}}{z}~}
\\[5mm] & =
\ln\pars{{\Gamma\pars{5/6}\Gamma\pars{1} \over \Gamma\pars{1/3}\Gamma\pars{1/2}}\,
{\Gamma\pars{1/6}\Gamma\pars{1/3} \over \Gamma\pars{2/3}\Gamma\pars{5/6}}}
\\[5mm] & =
\ln\pars{{1 \over \Gamma\pars{1/2}}\,
{\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}\qquad
\pars{\ds{\mbox{Note that}\ \Gamma\pars{1} = 1\ \mbox{and}\
\Gamma\pars{1 \over 2} = \root{\pi}}}
\\[5mm] & = \bbx{\ln\pars{{1 \over \root{\pi}}\,
{\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}} \approx 0.8412
\end{align}
Note that
$\ds{\left.\vphantom{\Large A}\Psi\pars{z}\right\vert_{\ \Re\pars{z}\ >\ 0} =
-\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant. See $\ds{\color{#000}{\mathbf{6.3.22}}}$ in A & S Table.
$$J(a)=\int_0^{\infty}dx\frac{e^{at}(e^{2x}-1)}{(e^{3x}+1) x}$$
Now consider a derivative with respect to $a$ we get (after rescaling) $$ 3J'a=\int_0^{\infty}\frac{e^{ (a+2)x}-e^{a x}}{e^x+1}=I'{a/3+2/3}-I'_{a/3} $$
expanding the denominator as geometric series and integrating termwise yields ...
– tired Jul 16 '17 at 13:10and likewise $$ I'{a/3+2/3}=\sum{n\geq0}\frac{(-1)^n}{n+a/3}=\psi(\frac 13-\frac a6)-\psi(\frac16-\frac a6) $$
where $\psi(z) =\log(\Gamma(z))'$ is the digamma function the digamma function. Integrating with respect to $a$ therefore yields
$$6J_a=\log(\Gamma(1-\frac a6))-\log(\Gamma(\frac12-\frac a6))-\log(\Gamma(\frac13-\frac a6))+\log(\Gamma(\frac16-\frac a6))+C$$
Now your integral in question is just $J_1$...
– tired Jul 16 '17 at 13:11int -(1-x)/((1-x+x^2)log(x)) dx, from x =0 to x=1to replace $$\int_0^1 -\frac{(1-x)}{(1-x + x^2)\log(x)} = \int_0^1 \frac{x-1}{(x^2 -x+1)\log(x)}$$ – amWhy Jul 16 '17 at 16:13