Evaluating $\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx$ This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$
where  $\arctan{x}=\tan^{-1}{x}$
mark: this integral is my favorite one. Thanks to whoever has nice methods.
I have proved the following:
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$
where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
I also have got a few by-products 
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$
where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$
and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$
and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant. 
 A: Splitting the integral at $x=0$, letting $x\mapsto -x$ in the first integral and then adding it to the second one, we arrive at
$$\int_{-1}^1\frac{\arctan(x)\log((1+x^2)/2)}{1+x}\textrm{d}x=2\int_0^1 \frac{x\arctan(x) \log(2/(1+x^2))}{1-x^2}\textrm{d}x=\frac{\pi^3}{96},$$
where the last integral is extracted in this solution magically exploiting the symmetry of a triple integral.
End of story
A: FWIW, here's Maple:
> f:= arctan(x)/(1+x)*ln((1+x^2)/2);
> int(f, x=-1..1);

$$ {\frac {7}{64}}\,{\pi }^{3}-{\frac {5}{16}}\,\pi \, \left( \ln 
 \left( 2 \right)  \right) ^{2}-\ln  \left( 2 \right) {\it Catalan}+1/
2\, \left( \ln  \left( 1-i \right)  \right) ^{2}\pi -1/2\,i \left( 
\ln  \left( 1+i \right)  \right) ^{2}\ln  \left( 2 \right) +\ln 
 \left( 1+i \right) {\it Catalan}+1/2\, \left( \ln  \left( 1+i
 \right)  \right) ^{2}\pi -i{\it polylog} \left( 3,-i \right) +i{\it 
polylog} \left( 3,i \right) +1/2\,i \left( \ln  \left( 1-i \right) 
 \right) ^{2}\ln  \left( 2 \right) -1/2\,i \left( \ln  \left( 1-i
 \right)  \right) ^{3}-1/48\,i\ln  \left( 1+i \right) {\pi }^{2}+1/2\,
i \left( \ln  \left( 1+i \right)  \right) ^{3}+\ln  \left( 1-i
 \right) {\it Catalan}+1/48\,i\ln  \left( 1-i \right) {\pi }^{2}
$$
> simplify(%);

$$ \frac{\pi^3}{96}
$$
I don't know if that qualifies as "nice", but it's certainly easy.
A: Here is a solution that only uses complex analysis:
Let $\epsilon$ > 0 and consider the truncated integral
$$ I_{\epsilon} = \int_{-1+\epsilon}^{1} \frac{\arctan x}{x+1} \log\left( \frac{1+x^2}{2} \right) \, dx. $$
By using the formula
$$ \arctan x = \frac{1}{2i} \log \left( \frac{1 + ix}{1 - ix} \right) = \frac{1}{2i} \left\{ \log \left( \frac{1+ix}{\sqrt{2}} \right) - \log \left( \frac{1-ix}{\sqrt{2}} \right) \right\}, $$
it follows that
$$ I_{\epsilon} = \Im \int_{-1+\epsilon}^{1} \frac{1}{x+1} \log^{2} \left( \frac{1+ix}{\sqrt{2}} \right) \, dx. $$
Now let $\omega = e^{i\pi/4}$ and make the change of variable $z = \frac{1+ix}{\sqrt{2}}$ to obtain
$$ I_{\epsilon} = \Im \int_{L_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz, $$
where $L_{\epsilon}$ is the line segment joining from $\bar{\omega}_{\epsilon} := \bar{\omega} + \frac{i\epsilon}{\sqrt{2}}$ to $\omega$. Now we tweak this contour of integration according to the following picture:

That is, we first draw a clockwise circular arc $\gamma_{\epsilon}$ centered at $\bar{\omega}$ joining from $\bar{\omega}_{\epsilon}$ to some points on the unit circle, and draw a counter-clockwise circular arc $\Gamma_{\epsilon}$ joining from the endpoint of $\gamma_{\epsilon}$ to $\omega$. Then
$$ I_{\epsilon} = \Im \int_{\gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz + \Im \int_{\Gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz =: J_{\epsilon} + K_{\epsilon}. $$
It is easy to check that as $\epsilon \to 0^{+}$, the central angle of $\gamma_{\epsilon}$ converges to $\pi / 4$. Since $\gamma_{\epsilon}$ winds $\bar{\omega}$ clockwise, we have
$$ \lim_{\epsilon \to 0^{+}} J_{\epsilon} = \Im \left( -\frac{i \pi}{4} \mathrm{Res}_{z=\bar{\omega}} \frac{\log^2 z}{z - \bar{\omega}} \right) = \frac{3}{2} \frac{\pi^3}{96}. $$
Also, by applying the change of variable $z = e^{i\theta}$,
$$ K_{\epsilon} = -\Re \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{1 - \bar{\omega}e^{-i\theta}} \, d\theta = \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{2} \, d\theta. $$
Thus taking $\epsilon \to 0^{+}$, we have
$$ \lim_{\epsilon \to 0^{+}} K_{\epsilon} = - \int_{0}^{\frac{\pi}{4}} \theta^2 \, d\theta = - \frac{1}{2} \frac{\pi^3}{96}. $$
Combining these results, we have
$$ \int_{-1}^{1} \frac{\arctan x}{x+1} \log \left( \frac{x^2 + 1}{2} \right) \, dx = \frac{\pi^3}{96}. $$
The same technique shows that
$$ \int_{-1}^{1} \frac{\arctan (t x)}{x+1} \log \left( \frac{1 + x^2 t^2}{1 + t^2} \right) \, dx = \frac{2}{3} \arctan^{3} t, \quad t \in \Bbb{R} .$$
