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
Unfortunately, standard integration techniques will not help you solve this integral. The actual anti-derivative of this function is huge (according to Wolfram alpha at least, see: https://www.wolframalpha.com/input/?i=integral+arctan%281%2F%28cosh%28x%29%29%29 ). To combat this, will we use a method called Feynman Integration (named after the Physicist Richard Feynman. Although the actual rule was discovered by Leibniz - who independently discovered Calculus).
Let
$${I(t)=\int_{0}^{\infty}\arctan\left(\frac{t}{\cosh(x)}\right)dx}$$
So we have defined a function in terms of our integral. Using the Leibniz rule for integration we get
$${I'(t)=\int_{0}^{\infty}\frac{\text{sech}(x)}{t^2\text{sech}^2(x) + 1}dx}$$
(to take the derivative, you take take the partial derivative of the inside :D). The inner function now has an elementary anti-derivative; namely
$${\int\frac{\text{sech}(x)}{1+t^2\text{sech}^2(x)}dx=\frac{-\arctan\left(\sqrt{t^2 + 1}\text{csch}(x)\right)}{\sqrt{1+t^2}} + C}$$
Hence the integral for ${I'(t)}$ can be found by taking limits:
$${\int_{0}^{\infty}\frac{\text{sech}(x)}{t^2\text{sech}^2(x) + 1}dx=\lim_{x\rightarrow \infty}\frac{-\arctan\left(\sqrt{t^2 + 1}\text{csch}(x)\right)}{\sqrt{1+t^2}} - \lim_{x\rightarrow 0}\frac{-\arctan\left(\sqrt{t^2 + 1}\text{csch}(x)\right)}{\sqrt{1+t^2}}}$$
$${\Rightarrow I'(t) = \frac{\pi}{2}\frac{1}{\sqrt{1+t^2}}}$$
So to find ${I(t)}$ we now simply integrate with respect to ${t}$ and find the constant. This gives us
$${I(t)=\frac{\pi}{2}\int\frac{1}{\sqrt{t^2 + 1}}dt=\frac{\pi}{2}\sinh^{-1}(t) + C}$$
(${\int\frac{1}{\sqrt{1+t^2}}dt}$ is just a known integral).
But ${I(0)=0\Rightarrow C=0}$ (since ${\sinh^{-1}(0)=0}$), hence
$${\int_{0}^{\infty}\arctan\left(\frac{n}{\cosh(x)}\right)dx=\frac{\pi}{2}\sinh^{-1}(n)}$$, but
$${\sinh^{-1}(n)=\ln\left(\sqrt{n^2 + 1} + n\right)}$$
and so indeed
$${\int_{0}^{\infty}\arctan\left(\frac{n}{\cosh(x)}\right)dx=\frac{\pi}{2}\ln\left(\sqrt{n^2 + 1} + n\right)}$$