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*}
We find an explicit closed form for all $n\in \mathbb{N}$. Using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and the geometric series (in the distributional sense, e.g. see the last remark of this answer) we obtain
$$\int_0^{\pi/2} x^n \csc(x)~dx=2i \int_0^{\pi/2} x^n \frac{e^{-ix}}{1-e^{-2ix}}~dx=2i\sum_{k=0}^{\infty} \int_0^{\pi/2} x^n e^{-(2k+1)ix}~dx. \tag{1}$$
We have the indefinite integral (eq. 2.321 in Gradshteyn and Ryzhik 7th ed.) for any $a\in \mathbb{C}$
$$\int x^n e^{ax}~dx=e^{ax}\left(\sum_{j=0}^n \frac{(-1)^j j! \binom{n}{j}}{a^{j+1}}x^{n-j}\right)+C,$$
which results in the definite integral
$$\int_0^{\pi/2} x^n e^{ax}~dx=e^{a\pi/2} \left(\sum_{j=0}^{n} \frac{(-1)^j j! \binom{n}{j}}{a^{j+1}}\left(\frac{\pi}{2}\right)^{n-j}\right)-\frac{(-1)^n n!}{a^{n+1}}.$$
Hence for $a=-(2k+1)i$ we have
$$\int_0^{\pi/2} x^n e^{-(2k+1)ix}~dx=i\cdot (-1)^k \left(\sum_{j=0}^{n} \frac{j! \binom{n}{j}}{i^{j+1} (2k+1)^{j+1}}\left(\frac{\pi}{2}\right)^{n-j}\right)+\frac{n!}{i^{n+1} (2k+1)^{n+1}}.$$
With this result, one can express $(1)$ in terms of Dirichlet beta functions
$$\beta(s):=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^s},$$
and Dirichlet lambda functions
$$\lambda(s):=\sum_{k=0}^{\infty} \frac{1}{(2k+1)^s}=(1-2^{-s})\zeta(s),$$
where the final equality is valid for all $\Re(s)>1$.
Thus, the integral is given by the following if $n$ is even (we use that $i^l=i^{-l}$ if $l$ is even):
$$\scriptsize \begin{align*}\int_0^{\pi/2} x^n \csc(x)~dx&=2 i^n\cdot n!\lambda(n+1)-2 \sum\limits_{\substack{0\leq j\leq n \\ j~\mathrm{odd}}} \left(\frac{\pi}{2}\right)^{n-j} i^{j+1} j! \binom{n}{j}\beta(j+1)\\&=\frac{(-1)^{n/2}\cdot n! (2^{n+1}-1)}{2^n}\zeta(n+1)-2 \sum\limits_{\substack{0\leq j\leq n \\ j~\mathrm{odd}}} \left(\frac{\pi}{2}\right)^{n-j} i^{j+1} j! \binom{n}{j}\beta(j+1)\\&=\frac{(-1)^{n/2} \cdot n! (2^{n+1}-1)}{2^n}\zeta(n+1)+2\sum_{m=1}^{\lfloor (n+1)/2\rfloor} \left(\frac{\pi}{2}\right)^{n+1-2m} (-1)^{m-1} (2m-1)! \binom{n}{2m-1}\beta(2m). \end{align*}$$
Otherwise, if $n$ is odd
$$\begin{align*}\int_0^{\pi/2} x^n \csc(x)~dx&=-2 \sum\limits_{\substack{0\leq j\leq n \\ j~\mathrm{odd}}} \left(\frac{\pi}{2}\right)^{n-j} i^{j+1} j! \binom{n}{j}\beta(j+1)\\&=2\sum_{m=1}^{\lfloor (n+1)/2\rfloor} \left(\frac{\pi}{2}\right)^{n+1-2m} (-1)^{m-1} (2m-1)! \binom{n}{2m-1}\beta(2m). \end{align*}$$
This agrees with @metamorphy's results (and yours). Note that we only sum over odd $j$ since the summands for even $j$ only contribute to the imaginary part of the integral, which we know is $0$ since the integrand is real on the domain of integration.
Best Answer
The general "closed-form" is: $$\int _0^1\frac{\arctan \left(x\right)\ln ^{2n}\left(x\right)}{1+x}\:dx$$ $$=\frac{\pi }{4}\left(1-2^{-2n}\right)\zeta \left(2n+1\right)\left(2n\right)!+\frac{1}{2}\beta \left(2n+2\right)\left(2n\right)!-\frac{\pi }{16}$$ $$\lim _{s\to 0}\left(\frac{d^{2n}}{ds^{2n}}\left(\csc \left(\frac{\pi s}{2}\right)\left(\psi \left(\frac{3-s}{4}\right)-\psi \left(\frac{1-s}{4}\right)\right)+\sec \left(\frac{\pi s}{2}\right)\left(\psi \left(1-\frac{s}{4}\right)-\psi \left(\frac{1}{2}-\frac{s}{4}\right)\right)-2\pi \csc \left(\pi s\right)\right)\right)$$ And you may find its evaluation in the book (Almost) Impossible Integrals, Sums, and Series through pages $140-145$ where double integration, symmetry and the following result are heavily exploited: $$\int _0^{\infty }\frac{x^s}{\left(1+x\right)\left(1+y^2x^2\right)}\:dx=\frac{\pi }{2}\csc \left(\frac{\pi s}{2}\right)\frac{y^{-s}}{1+y^2}+\frac{\pi }{2}\sec \left(\frac{\pi s}{2}\right)\frac{y^{1-s}}{1+y^2}-\frac{\pi \csc \left(\pi s\right)}{1+y^2}$$