How can one show that $\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?$ How can one show that 

$$\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?\tag1$$

I have tried IBP but it is seem too complicate. I am not sure what to do or how to tackle this problem. Please any help? Thank!
 A: Enforcing a substitution of $x \mapsto \sqrt{x}$ the integral becomes
$$I = \frac{1}{4} \int_0^1 \frac{\ln x \ln (1 - x)}{\sqrt{1 - x}} \, dx.$$
From the definition for Euler's beta function, namely
$$\text{B}(x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt,$$
we observe that
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \int_0^1 \frac{\ln t \ln (1 - t)}{\sqrt{1 - t}} \, dt.$$
On finding the required derivatives for the Beta function we obtain
$$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [\{\psi(x) - \psi(x + y)\}\{\psi(y) - \psi (x + y)\} - \psi^{(1)}(x + y) \right ].$$
Here $\psi(x)$ is the digamma function while $\psi^{(1)}(x)$ is the trigamma function or polygamma function of order one. 
On taking the required limits we are left with
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \text{B} \left (1, \frac{1}{2} \right ) \left [\{\psi(1) - \psi(3/2)\}\{\psi (1/2) - \psi(3/2)\} - \psi^{(1)}(3/2) \right ].$$
Each of the values for the digamma and polygamma functions are well known. They are
\begin{align*}
\psi \left (\frac{1}{2} \right ) = -\gamma - \ln (4), & \quad \psi (1) = -\gamma,\\
\psi \left (\frac{3}{2} \right ) = 2 - \gamma - \ln (4),& \quad 
\psi^{(1)} \left (\frac{3}{2} \right ) = 3 \zeta (2) - 4 = \frac{\pi^2}{2} - 4.
\end{align*}
As
$$\text{B} \left (1, \frac{1}{2} \right ) = \frac{\Gamma (1) \Gamma (1/2)}{\Gamma (3/2)} = 2,$$
we have
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = 16 - 4 \ln (4) - \pi^2,$$
yielding
$$\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx = 4 - \ln (4) - \frac{\pi^2}{4}.$$
A: Let $x=\sin t$ and then
$$I=\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx =2\int_0^{\pi/2}\sin t\ln\sin t\ln\cos tdt=-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t.$$
Letting $u=\cos t$ gives
\begin{eqnarray}
I&=&-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t\\
&=&\int_0^1\ln(1-u^2)\ln udu\\
&=&-\int_0^1\sum_{k=1}^\infty\frac1{k}u^{2k}\ln udu\\
&=&\sum_{k=1}^\infty\frac1{k(2k+1)^2}\\
&=&\sum_{k=1}^\infty\left[\frac{1}{k(2k+1)}-\frac{2}{(2k+1)^2}\right].
\end{eqnarray}
Now using
$$　\sum_{k=1}^\infty\frac{1}{k(2k+1)}=2-2\ln2 $$
and
$$ \sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8} $$
it is easy to get
$$ I=4-2\ln2-\frac{\pi^2}{4}.$$
