$$\int_{0}^{1} \log\left(\frac{(\sqrt{1 - x} - 1)(\sqrt{1 - x^2} + 1)}{(\sqrt{1 - x} + 1)(\sqrt{1 - x^2} - 1)}\right) \arcsin(x^2) \, dx = \frac{\pi^3}{12} - \pi\left(6\sqrt{2} + 2 + \log\left(\frac{3}{4} - \frac{1}{\sqrt{2}}\right)\right) - 4(2C - 7)$$
$$= \int_{0}^{1} \log\left(\frac{(1 - \sqrt{1 - x})(1 + \sqrt{1 - x^2})}{(1 + \sqrt{1 - x})(1 - \sqrt{1 - x^2})}\right) \arcsin(x^2) \, dx$$
$$=A+B$$ where
$$ A = \int_{0}^{1} \log\left(\frac{1 - \sqrt{1 - x}}{1 + \sqrt{1 - x}}\right) \arcsin(x^2) \, dx $$ $$ B = \int_{0}^{1} \log\left(\frac{1 + \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}}\right) \arcsin(x^2) \, dx $$
\begin{aligned} B &\stackrel{x = \sin t}{=} -2 \int_{0}^{\frac{\pi}{2}} t^2 \log\left(\tan\left(\frac{t}{2}\right)\right) \cos t \, dt \\ &= -16 \int_{0}^{\frac{\pi}{4}} t^2 \log(\tan t) \cos 2t \, dt = \ldots \end{aligned}