Hard Integral $\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$ I am trying to calculate the integral $$\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$$
My idea is to use and try to calculate and prove that
$\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=-\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}-\int\limits_{0}^{1}{\frac{2x\arctan x}{1+x^{2}}\ln \left( 1-x^{2} \right)dx}$
$\frac{2x\arctan x}{1+x^{2}}=\operatorname{Re}\left( \frac{\ln \left( 1+ix \right)}{1+ix} \right)-\operatorname{Re}\left( \frac{\ln \left( 1-ix \right)}{1+ix} \right)$
$\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{32}+\frac{\pi }{4}\ln ^{2}2-2C\ln 2$
$\int\limits_{0}^{1}{\frac{\ln \left( 1\pm ix \right)\ln \left( 1\pm x \right)}{1+ix}dx}$
 A: It seems the integral is designed to intentionally benefit from the following facts:
$$\log \frac{1+x^{2}}{2}=\log(\frac{1}{2}(x+i)(x-i))=\log(\frac{x+i}{\sqrt 2})+\log(\frac{x-i}{\sqrt 2})$$ 
and 
$$\arctan (x)= \frac{1}{2}i\log(1-i x)-\frac{1}{2}i\log(1+i x)= \frac{1}{2i}(\log(\frac{x+i}{\sqrt 2})-\log(\frac{x-i}{\sqrt 2}))$$ 
Let's work on the anti derivative:
$$\int{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=-\frac{1}{2i}\int{\frac{1}{1+x}(\log^2(\frac{x+i}{\sqrt 2})-\log^2(\frac{x-i}{\sqrt 2}))dx}=\text{Im}{\int{\frac{1}{1+x}\log^2(\frac{x-i}{-i\sqrt 2})dx}}=\text{Im}{\int{\frac{1}{1+x}\log^2(\frac{1+ix}{\sqrt 2})dx}}$$
Encapsulating $y=\frac{1+ix}{\sqrt 2}$ (i.e. $1+x=1+(1-y\sqrt 2)i$ and $dx=-idy\sqrt{2}$), the anti derivative can be simplified to
$$\int{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\text{Im}{\int{-\frac{\log^2y}{1+(1-y\sqrt 2)i}i\sqrt{2}dy}}=\text{Im}{\int{\frac{\log^2y}{y+\frac{i-1}{\sqrt{2}}}dy}}$$
Using dilogarithm function and assuming $k=-\frac{i-1}{\sqrt{2}}$, yields
$$\int \frac{\log^2y}{y-k}dy=-2\text{Li}_3(ky)+2\text{Li}_2(ky)\log(y)+\log(1-ky)\log^2(y)+\text{constant}$$
Now, computing the finite integral should be doable. wolframalpha gives me $\pi^3/96$ as it is stated in the question. I won't be surprised if there are typos or errors - you're welcome to fix or improve.
