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
$$ \int_{-1}^{1} \frac{\arctan (t x)}{x+1} \log \left( \frac{1 + x^2 t^2}{1 + t^2} \right) \, dx = \frac{2}{3} \arctan^{3} t, \quad t \in \Bbb{R} .$$
I tried 2 ways to find a closed form, though unsuccessful up to this point.
1st trial. Let $I$ denote the integral, and write
$$ I = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-(2n+1)x})}{x (1 + e^{-x})^{2}} \, dx. $$
In order to evaluate the integral inside the summation, we introduce new functions $I(s)$ and $J_n(s)$ by
$$ I(s) = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{x^{s-1} e^{-x}(1 - e^{-(2n+1)x})}{(1 + e^{-x})^{2}} \, dx =: 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} J_n(s) $$
so that $I = I(0)$. Then it is easy to calculate that for $\Re(s) > 1$, $J_n(s)$ is written as
$$ J_n(s) = \Gamma(s) \left( \eta(s-1) + \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s-1}} - (2n+1) \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s}} \right), $$
where $\eta$ is the Dirichlet eta function. Plugging this back to $I(s)$ and manipulating a little bit, we obtain
$$ I(s) = 8\Gamma(s) \left( \frac{\pi}{4} \eta(s-1) - 4^{-s}\left( \zeta(s, \tfrac{1}{4}) - \zeta(s, \tfrac{1}{2}) \right) + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=2n+1}^{\infty} \frac{(-1)^{k-1}}{k^{s-1}} \right). $$
This is valid for $\Re(s) > 1$. But if we can somehow manage to find an analytic continuation of the last summation part, then we may find the value of $I = I(0)$.
2nd trial. I began with the following representation
\begin{align*}
I
&= -2 \int_{0}^{\infty} \frac{1-e^{-t}}{1+e^{-t}} \left( \frac{1}{\cosh t} - \frac{2}{t} ( \arctan(1) - \arctan (e^{-t})) \right) \, \frac{dt}{t} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)(2n+2)} \int_{-\infty}^{\infty} \frac{\tanh^{2(n+1)} x}{x^{2}} \, dx.
\end{align*}
With some residue calculation, we can find that
\begin{align*}
\int_{-\infty}^{\infty} \frac{\tanh^{2n} x}{x^{2}} \, dx
&= \frac{2}{i\pi} \, \underset{z=0}{\mathrm{Res}} \left[ \psi_{1}\left(\tfrac{1}{2} + \tfrac{1}{i\pi} z\right) \coth^{2n} z \right] \\
&= 2^{2n+3} \sum_{m=1}^{n} (-1)^{m-1}m (1-2^{-2m-1}) A_{n-m}^{(2n)} \, \frac{\zeta(2m+1)}{\pi^{2m}},
\end{align*}
where $A_m^{(n)}$ is defined by the following combinatoric sum
$$ A_m^{(n)} = \sum_{\substack{ k_1 + \cdots + k_n = m \\ k_1, \cdots, k_n \geq 0 }} \frac{B_{2k_1} \cdots B_{2k_n}}{(2k_1)! \cdots (2k_n)!} = 2^{-2m} [z^{2m}](z \coth z)^{n} \in \Bbb{Q}, $$
where $B_k$ are Bernoulli numbers. Still the final output is egregiously complicated, so I stopped here.
3rd trial. The following yet another representation may be helpful, I guess.
$$ I = \int_{0}^{1/2} \frac{1 - \cot(\pi u/2)}{2} \left\{ \psi_1\left(\tfrac{1+u}{2}\right) - \psi_1\left(\tfrac{1-u}{2}\right) \right\} \, du. $$
Best Answer
\begin{align}J&=\int_0^1 \frac{x\arctan x}{1+x^4}dx\\ &=\int_0^1 \frac{x}{1+x^4}\int_0^1 \left(\frac{x}{1+t^2x^2}dt\right)dx\\ &=2\left(\int_0^1 \frac{x^2}{1+x^4}dx\right)\left(\int_0^1 \frac{1}{1+t^4}dt\right)-\int_0^1\frac{t}{1+t^4}\int_0^1 \left(\frac{t}{1+t^2x^2}dx\right)dt\\ &=2\left(\underbrace{\int_0^1 \frac{x^2}{1+x^4}dx}_{=B}\right)\left(\underbrace{\int_0^1 \frac{1}{1+t^4}dt}_{=A}\right)-J\\ J&=\boxed{AB}\\ K&=\int_0^1 \frac{x^3\text{arctanh }x}{1+x^4}dx\\ &=\int_0^1 \frac{x^3}{1+x^4}\left(\int_0^1 \frac{x}{1-t^2x^2}dt\right)dx\\ &=-A^2-B^2+\int_0^1 \frac{1}{t(1+t^4)}\left(\int_0^1 \frac{t}{1-t^2x^2}dx\right)dt\\ &=-A^2-B^2+\underbrace{\int_0^1 \frac{\text{arctanh }t}{t}dt}_{u=\frac{1-t}{1+t}}-K\\ K&=-\frac{A^2}{2}-\frac{B^2}{2}-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln u}{1-u^2}du}_{=-\frac{\pi^2}{8}}\\ &=\boxed{-\frac{A^2}{2}-\frac{B^2}{2}+\frac{\pi^2}{16}} \end{align} On the other hand, \begin{align}2J-2K&=2AB+A^2+B^2-\frac{\pi^2}{8}\\ &=(A+B)^2-\frac{\pi^2}{8}\\ &=\left(\underbrace{\int_0^1 \frac{1+x^2}{1+x^4}dx}_{=Z}\right)^2-\frac{\pi^2}{8}\\ Z&=\int_0^1 \frac{1+\frac{1}{x^2}}{\left(\frac{1}{x}-x\right)^2+2}dx\\ &\overset{u=\frac{1}{x}-x}=\int_0^\infty \frac{1}{2+u^2}du\\ &=\frac{\pi }{2\sqrt{2}}\\ \end{align} Therefore, \begin{align}Z^2&=\frac{\pi^2}{8}\\ 2J-2K&=0\\ &\boxed{J=K} \end{align}