An integration $ \int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x.$ 
How can we evaluate 
  $$\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x?$$

Usually when having as the integrand a logarithmic function, the first thing would be to try to integrate by parts, however in this case there's a product of logarithms and trying to integrate by parts would just produce two harder integrals and it might not be a good approach.
There is also An integral by O. Furdui $\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$ which is quite similar to this one, however the substitution $\sqrt{1+x}-\sqrt{1-x}=2\sin t$ doesn't seem to be useful due to the $\sqrt{1+x}+\sqrt{1-x}$ term. 
Is there a better approach to this integral that reduces it to something simpler?

One may prove that

$$
\int_0^1 x\ln \left( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left( \sqrt {1+x} -\sqrt{1-x}\right)\:\mathrm{d}x=\frac{\ln^2 2}4-\frac{3\ln2}4+\frac1{16}. \tag1
$$

Hint. By using
$$
\begin{align}
4ab&=(a+b)^2-(a-b)^2
\end{align}
$$ giving
$$
\begin{align}
4\ln b \ln c&=\ln^2 (bc)-\ln^2 \left(\frac bc\right)
\end{align}
$$ and by observing that
$$
\begin{align}
&\left ( \sqrt{1+x}+\sqrt{1-x}\right)\left ( \sqrt {1+x} -\sqrt{1-x} \right)=2x
\\
&\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt {1+x} +\sqrt{1-x}}=\frac{x}{1+\sqrt{1-x^2}},\quad 0<x<1,
\end{align}
$$ one gets
$$
\begin{align}
4&\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\:\mathrm{d}x\\=&\int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x
-\int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:\mathrm{d}x. \tag2
\end{align}
$$ Then integrating by parts twice one gets
$$
\int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x=\frac{\ln^2 2}2-\frac{\ln2}2+\frac14, \tag3
$$ by the change of variable $u=\dfrac{x}{1+\sqrt{1-x^2}}$, one has
$$
\begin{align}
\int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:dx&=\int_0^1\frac{4u(1-u^2)}{(1+u^2)^3}\cdot \ln^2 u\:du
\\&=\frac12\int_0^1\frac{(1-v)}{(1+v)^3}\cdot \ln^2 v\:du
\\&= \ln 2,\tag4\end{align}
$$ obtained by parts. By inserting $(4)$ and $(3)$ into $(2)$ one gets $(1)$.
