A long integral...
We are to evaluate the following integral, \begin{equation} \boxed{ I = \int_0^1 x \log(\sqrt{1+x}+\sqrt{1-x})\log(\sqrt{1+x}-\sqrt{1-x})dx } \label{integral} \tag{I} \end{equation} Define the following two functions: \begin{subequations} \begin{align} p(x)&=\sqrt{1+x}+\sqrt{1-x} \label{plus}\\ m(x)&=\sqrt{1+x}-\sqrt{1-x} \label{minus} \end{align} \end{subequations} The product of these two functions is, \begin{equation} p(x)m(x)=2x \Rightarrow x =\frac{1}{2}p(x)m(x) \label{product} \end{equation} (\ref{integral}), (\ref{plus}), (\ref{minus}) and (\ref{product}) can be combined to give, \begin{equation} I = \frac{1}{2}\int_0^1 p(x)m(x) \log p(x)\log m(x)dx \label{allispm} \end{equation} Since, \begin{equation} \frac{dm(x)}{dx}=\frac{1}{2}\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1-x}}\right) =\frac{1}{2} \frac{p(x)}{\sqrt{1-x^2}} \label{mderiv} \end{equation} we can combine (\ref{allispm}) and (\ref{mderiv})...