This approach is similar to Ron Gordon's answer. We introduce a new variable $\beta$ before differentiating.
By substitution $\tan x=u$,
$$\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\! \alpha \tan x\right)\, \mathrm{d}x=\int_{0}^{\infty}\frac{\tan^{-1}u\tan^{-1}\alpha u}{u^2}\,\mathrm{d}u$$
We will prove $$\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac\pi2\log\left(\frac{(\alpha+\beta)^{\alpha+\beta}}{\alpha^\alpha\beta^\beta}\right)\tag{1}$$
Differentiation gives
\begin{align}\partial_\alpha\partial_\beta\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u&=\int_{0}^{\infty}\frac{\mathrm{d}u}{(1+\alpha^2 u^2)(1+\beta^2 u^2)}\\&=\frac1{\alpha^2-\beta^2}\int_{0}^{\infty}\frac{\alpha^2}{1+\alpha^2u^2}-\frac{\beta^2}{1+\beta^2u^2}\,\mathrm{d}u\\&=\frac1{\alpha^2-\beta^2}\int_{0}^{\infty}\frac{\alpha}{1+u^2}-\frac{\beta}{1+u^2}\,\mathrm{d}u\\&=\frac\pi{2(\alpha+\beta)}\end{align}
for $\alpha\ne\beta$. The case $\alpha=\beta$ can be proved by letting $\alpha\to\beta$. Then
$$\partial_\beta\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac{\pi}{2}(\log(\alpha+\beta)-C_1(\beta))$$ If $\alpha\to0$ we have $0=\frac\pi2(\log\beta-C_1(\beta))$ so $C_1(\beta)=\log\beta$. Integrating by $\beta$,$$\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac{\pi}{2}((\alpha+\beta)\log(\alpha+\beta)-\beta\log\beta-C_2(\alpha))$$ Let $\beta\to0$ and we find $C_2(\alpha)=\alpha\log\alpha$. Thus $(1)$ is proven and putting $\beta=1$ gives the conclusion $$\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\! \alpha \tan x\right)\, \mathrm{d}x=\frac\pi2\log\left(\frac{(\alpha+1)^{\alpha+1}}{\alpha^\alpha}\right)$$
Here is another solution: Let $(I_n)$ by
$$ I_n = \int_{0}^{\infty} \frac{\arctan x}{x(1+x^2)^n} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sin\theta} \cos^{2n-1}\theta \, d\theta. $$
Then by a simple calculation,
$$ I_n - I_{n+1} = \int_{0}^{\frac{\pi}{2}} \theta \sin\theta \cos^{2n-1}\theta \, d\theta = \frac{1}{2n} \int_{0}^{\frac{\pi}{2}} \cos^{2n}\theta \, d\theta. $$
Since $I_n \to 0$ as $n \to \infty$, we find that
$$ I_n = \sum_{k=n}^{\infty} \frac{1}{2k} \int_{0}^{\frac{\pi}{2}} \cos^{2k}\theta \, d\theta. $$
Splitting the summation as $\sum_{k=n}^{\infty} = \sum_{k=1}^{\infty} - \sum_{k=1}^{n-1}$, we find that
$$ I_n = \frac{\pi}{2}\left( \log 2 - \sum_{k=1}^{n-1} \frac{1}{2k} \frac{(2k-1)!!}{(2k)!!} \right), $$
where $n!!$ denotes the double factorial.
Edit 1. In general, we have
$$ \int_{0}^{\infty} \frac{\arctan^s x}{x(1+x^2)^{n+1}} \, dx = \int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta - \sum_{k=1}^{n} \int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta. \tag{1} $$
Currently I have no idea how to obtain a simple formula for the following integral
$$ \int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta, \tag{2} $$
even when $s = 2$. On the other hand, for any $s > 0$ and $N \geq \lfloor s/2 \rfloor$ we have
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta
&= 2^{-s}\cos\left(\frac{\pi s}{2}\right)\Gamma(1+s)\zeta(1+s) \\
&\quad + \left(\frac{\pi}{2}\right)^s \sum_{k=0}^{N} (-1)^k \pi^{-2k} \frac{\Gamma(2k-s)}{\Gamma(-s)} \eta(2k+1) \\
&\quad + \frac{(-1)^{N+1}}{2^s \Gamma(-s)} \int_{0}^{\infty} \frac{t^{2N+1-s}}{1+t^2} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{1+s}} e^{-\pi n t} \right) \, dt,
\end{align*}
where $\eta(s)$ denotes the Dirichlet eta function. (My solution is somewhat involved, so I will post later if it seems useful to our problem.) In particular, when $s$ is a positive integer, then the integral part vanishes and the formula becomes much simpler. Thus the formula (*) gives a closed form as long as we can figure out the integral (2).
Example 1. For example, when $s = 2$ then we can use $N = 1$ and then
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \theta^2 \cot \theta \, d\theta
&= -\frac{1}{2}\zeta(3) + \frac{\pi^2}{4} \log 2 - \frac{1}{2}\eta(3) \\
&= \frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3).
\end{align*}
Since we can figure out the integral (2) for $s = 2$ and $k = 1, \cdots, 4$, we easily obtain OP's last identity.
Here is another example:
Example 2. Using the formula with $s = 6$, we can check that
\begin{align*}
\int_{0}^{\infty} \frac{\arctan^6 x}{x(1+x^2)^3} \, dx
&= \frac{\pi^6}{64} \log 2 -\frac{45 \pi^4}{128} \zeta(3) + \frac{675 \pi^2}{128} \zeta(5) -\frac{5715}{256} \zeta(7) \\
&\quad - \frac{11 \pi^6}{2048} + \frac{705 \pi^4}{4096} - \frac{8595 \pi^2}{4096} + \frac{135}{16} \\
&\approx 0.0349464822054751922142122595622\cdots.
\end{align*}
Best Answer
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
$$ \int_{-1}^{1} \frac{\arctan (t x)}{x+1} \log \left( \frac{1 + x^2 t^2}{1 + t^2} \right) \, dx = \frac{2}{3} \arctan^{3} t, \quad t \in \Bbb{R} .$$