How to evaluate $\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)dx?$ How may one demonstrate this integral is equal to this result?

$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)\mathrm dx=\pi\zeta(2)-2\pi[1-\ln^2(2)]$$

Where do I start from? 
 A: Just some thoughts, for the moment. Perhaps I'll add details later.
First of all, using the properties of logarithms you will easily find that your integral can be rewritten as
$$\int_0^1 \arccos\sqrt{1-x}\left(\ln^2(1-x) - \ln^2(x)\right)\ \text{d}x$$
A suitable change of variable like $1-x = z^2$ or some exponential manipulations shall work.
In any case you might also split the two integrals.
For the records, a very standard integral can be noticed right now:
$$\int_0^1 z \arccos(z) \ln^2(z)\ \text{d}z = \frac{1}{96} \pi  \left(\pi ^2-6+3 \ln ^2(4)\right)$$
Other manipulations may arise from Taylor series, but this is all to be understood properly.
A: The variable change and the integration by parts reveal together that
$$I= 8\underbrace{\int_0^{\pi/2}\sin^2(x)\log^2(\sin(x))\textrm{d}x}_{\displaystyle \text{Beta function}}-8\underbrace{\int_0^{\pi/2}\sin^2(x)\log(\sin(x))\textrm{d}x}_{\displaystyle \text{Beta function}}$$
$$=\pi \zeta(2)-2\pi (1-\log^2(2)).$$
Extra information: see eq. (14) here http://mathworld.wolfram.com/BetaFunction.html
A: This answer will primarily show how Waiting most likely got his answer. To start off, let's first define $I$ to be the integral under question, and using the basic logarithmic rules we all learned in first grade, $I$ can be transformed as
$$\begin{align*}I & =\int\limits_0^1dx\,\arccos\sqrt{1-x}\log^2(1-x)-\int\limits_0^1dx\,\arccos\sqrt{1-x}\log^2x\\ & =\int\limits_0^1dx\,\arccos\sqrt x\log^2x-\int\limits_0^1dx\,\arccos\sqrt x\log^2(1-x)\tag1\end{align*}$$
From ($1$), make the substitution $x\mapsto\sqrt x$ to get rid of the radical in the $\arccos(\cdot)$ function. Therefore, ($1$) turns into the form
$$\begin{align*}I & =8\int\limits_0^1dx\, x\arccos x\log^2x-2\int\limits_0^1dx\, x\arccos x\log^2(1-x^2)\tag2\end{align*}$$
Next, substitute $x\mapsto\cos x$ to get
$$I=8\int\limits_0^{\pi/2}dx\, x\sin x\cos x\log^2\cos x-8\int\limits_0^{\pi/2}dx\, x\sin x\cos x\log^2\sin x\tag3$$
Now, integration by parts, as difficult as it may seem, can be used to reduce the integral into something the beta function can easily manage. First, denote $I_1$ and $I_2$ as the first and second integral respectively of ($3$). Letting $u=x$ and integrating by parts on $I_1$ gives
$$\begin{align*}I_1 & =\frac {\pi}2+4\int\limits_0^{\pi/2}dx\,\cos^2x\log^2\cos x-4\int\limits_0^{\pi/2}dx\,\cos^2x\log\cos x\\ & =\frac {\pi}2+4\int\limits_0^{\pi/2}dx\,\sin^2\log^2\sin x-4\int\limits_0^{\pi/2}dx\,\sin^2x\log\sin x\\I_2 & =\frac {\pi}2-4\int\limits_0^{\pi/2}dx\,\sin^2x\log^2\sin x+4\int\limits_0^{\pi/2}dx\,\sin^2 x\log\sin x\end{align*}$$
Note that in the second line, we exploited the identity $\cos\left(\frac {\pi}2-x\right)=\sin x$ and used the limit identity for integrals. Putting everything together in ($3$) leaves
$$\begin{align*}I & =8\int\limits_0^{\pi/2}dx\,\sin^2\log^2\sin x-8\int\limits_0^{\pi/2}dx\,\sin^2x\log\sin x\\ & =8\frac {\partial^2}{\partial a^2}\int\limits_0^{\pi/2}dx\,\sin^{a+2}x-8\frac {\partial}{\partial b}\int\limits_0^{\pi/2}dx\,\sin^{b+2}x\end{align*}$$
Using equation ($14$) from here gives the answer as
$$\begin{align*}I & =4\sqrt{\pi}\frac {\partial^2}{\partial a^2}\frac {\Gamma\left(\frac a2+\frac 32\right)}{\Gamma\left(\frac a2+2\right)}-4\sqrt{\pi}\frac {\partial}{\partial b}\frac {\Gamma\left(\frac b2+\frac 32\right)}{\Gamma\left(\frac b2+2\right)}\\ & =\frac {\pi}6\biggr[\pi^2+6\log 2\left(\log 4-2\right)-6\biggr]-\pi+2\pi\log 2\end{align*}$$
A little simplifying reveals the answer
$$\int\limits_0^1dx\,\arccos\sqrt{1-x}\log(x-x^2)\log\left(\frac 1x-1\right)\color{blue}{=\frac {\pi^3}6+2\pi\log^22-2\pi}$$
