How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$ 
Evaluate
  $$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$  

I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got  
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$  
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.  
Please Help!

I dont'have (yet) a complete solution for now.
Let $I=\displaystyle \int_0^1 (\arctan(x))^2 \ln\Big(\dfrac{1+x^2}{2x^2}\Big)dx$
$I=\displaystyle \int_0^1 (\arctan(x))^2\ln(1+x^2)dx-2\int_0^1 (\arctan(x))^2\ln(x)dx-\ln(2)\int_0^1 (\arctan(x))^2dx$
I know how to compute the last one ;)
Let $J=\displaystyle \int_0^{\frac{\pi}{2}} \log(\sin x)dx$
The value of $J$ is well known to be $-\dfrac{\pi}{2}\ln 2$
Perform integration by parts:
$J=\displaystyle \Big[x\log(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx=-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx$
Let $K=-J$
Perform the change of variable $u=\tan x$
$K=\displaystyle \int_0^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
$K\displaystyle =\int_0^{1}\dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$
In the second integral, in the right member perform the change of variable $u=\dfrac{1}{x}$:
$K= \displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan \Big(\dfrac{1}{x}\Big) }{1+x^2}dx$
For $x>0$, $\arctan \Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
For $x\neq 0$ ,$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-2\int_0^1 \dfrac{x\arctan x }{1+x^2}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx$
The derivative of $(\arctan x)^2$ is $\dfrac{2\arctan x}{1+x^2}$
Hence:
$\displaystyle \int_0^1 \dfrac{2x\arctan x }{1+x^2}dx=\Big[x(\arctan x)^2\Big]_0^1-\int_0^1 (\arctan x)^2dx=\dfrac{\pi^2}{16}-\int_0^1 (\arctan x)^2dx$
Therefore:
$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1 (\arctan x)^2dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$
Recall $K=\dfrac{\pi}{2}\ln 2$
Therefore:
$\displaystyle\int_0^1 (\arctan x)^2 dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\ln 2$
Where $G=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx$ is the Catalan's constant.
