Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.0951875 ... $,
and its possible closed form is $ - \frac18\ln2\ln3$. But I don't know how to do it.
Thanks in advance.

Firstly, recognise the "symmetry" in the integral
$$I = \int_{0}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln \left(\frac{\ln(1/2+x)}{\ln(1/2-x)}\right)\, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln\left(-\ln\left(\frac{1}{2}+x\right)\right)\,dx$$
Now enforce the substitution $-\ln(1/2+x)=u$:
$$\implies I = \frac{1}{8} \int_{0}^{\infty} \frac{\ln(u) (2-e^u)}{e^{2u}-e^u+1}\,du$$
Now consider firstly the Mellin transform
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-z}\,dx = \frac{1}{z} \Gamma (s) \operatorname{Li}_s (z)$$
which can be found here. Now, through partial fractions, we can determine that
\begin{align}\int_{0}^{\infty} \frac{x^{s-1}}{(e^x+p-q)(e^x+p+q)}\,dx &= \int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}+2pe^x+p^2-q^2}\,dx\\&=\frac{\Gamma(s)}{2q(p+q)} \operatorname{Li}_s (-p-q) - \frac{\Gamma(s)}{2q(p-q)} \operatorname{Li}_s (-p+q)\end{align}
Thus setting $p=-\frac{1}{2}$ and $q=\frac{i \sqrt{3}}{2}$:
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}-e^x+1}\,dx=-\frac{\Gamma(s)}{2\sqrt{3}}\left(\left(\sqrt{3}-i\right)\operatorname{Li}_s\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)+\left(\sqrt{3}+i\right)\operatorname{Li}_s\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\right)$$
Through a similar process, we can determine $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{(e^x+p-q)(e^x+p+q)}\,dx = \frac{\Gamma(s)}{2q} \left(\operatorname{Li}_s (-p+q)-\operatorname{Li}_s (-p-q)\right)$$

In general:
$$\int_{0}^{\infty} \frac{x^{s-1}}{a \cosh x+b\sinh x+c}\,dx=\frac{\Gamma(s)}{\sqrt{b^2+c^2-a^2}} \left(\operatorname{Li}_s \left(\frac{\sqrt{b^2+c^2-a^2}-c}{a+b}\right)-\operatorname{Li}_s \left(-\frac{\sqrt{b^2+c^2-a^2}+c}{a+b}\right)\right)$$
which I utilised here.

This gives $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{e^{2x}-e^x+1} \,dx=-\frac{i\cdot\Gamma(s)}{\sqrt{3}}\left(\operatorname{Li}_s\left(\frac{1+i\sqrt{3}}{2}\right)-\operatorname{Li}_s\left(\frac{1-i\sqrt{3}}{2}\right)\right)$$
Thus $$\int_{0}^{\infty} \frac{x^{s-1}(2-e^x)}{e^{2x}-e^x+1}\,dx=-\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)$$
Giving finally
$$I = -\frac{1}{8}\cdot\frac{d}{ds}\left(\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)\right)\Bigg|_{s=1} = -\frac{1}{8} \ln(2) \ln(3)$$ as desired. $\square$
