I know how to prove that,
$\displaystyle K=\int_0^1 \frac{\ln^2(1-x^4)}{1+x^2}\,dx=\frac{9}{4}\pi\ln^2 2+\frac{7}{48}\pi^3-12G\ln 2$
(see: Closed form for the following integrals $ 1)\int_0^1\frac{\ln(1-x^2)\ln (1+x^2)}{1+x^2} \; dx$ )
But, i'm searching for an alternative proof using Beta function of Euler
My attempts:
$\displaystyle K=\int_0^1\left(\frac{\ln^2(1-x^4)}{1-x^4}-\frac{x^2\ln^2(1-x^4)}{1-x^4}\right)\,dx$
But,
$\displaystyle \int_0^1 \frac{\ln^2(1-x^4)}{1-x^4}\,dx$ and $\displaystyle \int_0^1 \frac{x^2\ln^2(1-x^4)}{1-x^4}\,dx$ do not converge.
Second attempt,
$\displaystyle K=\frac{1}{48}\int_0^1 \left(\sqrt{x}-3\right)x^{-\frac{7}{4}}\ln^3(1-x)\,dx$
But, for example, if $s\geq 0$,
$\displaystyle \int_0^1 x^{-\frac{7}{4}}(1-x)^s\,dx$
does not converge.
Third attempt,
$\displaystyle K=-\frac{1}{4}\int_0^1 \frac{\left(\sqrt{1-x}-1\right)\ln^2 x}{x\left(1-x\right)^{\frac{3}{4}}}\,dx$
But, $\displaystyle \int_0^1 \frac{\ln^2 x}{x\left(1-x\right)^{\frac{3}{4}}}\,dx$
does not converge.