Is there an exact solution for $\large\int \frac{dx}{\tan^{-1}(x)}$?

Let $$I(x)=\int\limits_1^x \dfrac{\text dt}{\arctan t},\tag1$$ then $$I(\tan y)=\int\limits_{\large^\pi/\mspace{1mu}_4}^y\dfrac1t\,\text d\tan t \;\overset{\text{IBP}}{=\!=}\;\dfrac{\tan t}t\bigg|_{\large^\pi/\mspace{1mu}_4}^y+\int\limits_{\large^\pi/\mspace{1mu}_4}^y\,\dfrac{\tan t}{t^2}\,\text dt = \dfrac{\tan y}y-\dfrac4\pi +\int\limits_{\large^\pi/\mspace{1mu}_4}^y\dfrac{\tan t}{t^2}\,\text dt,\tag2$$ or, applying the known Maclaurin series of the tangent function in the form of $$\tan t =\sum\limits_{n=1}^\infty (-1)^{n-1}\dfrac{4^n(4^n-1)\text B_{2n}}{(2n)!}t^{2n-1}= t+\dfrac13\,t^3+\dfrac2{15}\,t^5+\dfrac{17}{315}\,t^7+\dfrac{62}{2835}\,t^9+\dots,\tag3$$

$$\color{green}{\mathbf{I(\tan y)= \dfrac{\tan y}y-\dfrac4\pi+\ln\dfrac{4y}\pi + \sum\limits_{n=2}^\infty c_n \left(y^{2n-2}-\left(\dfrac\pi4\right)^{2n-2}\right),}}\tag4$$ where $$\color{green}{\mathbf{c_n = (-1)^{n-1}\dfrac{4^n(4^n-1)\text B_{2n}}{(2n-2)(2n)!}.}}\tag5$$

I.e., we have got $\;1D\;$ series.

Alternative form of the solution $(4)$ is $$\color{green}{\mathbf{I(x)= \dfrac x{\arctan x}-\dfrac4\pi+\ln\dfrac{4\arctan x}\pi + \sum\limits_{n=2}^\infty c_n\left(\arctan^{2n-2} x -\left(\dfrac\pi4\right)^{2n-2}\right).}}\tag6$$

It is known that $$ \csc^2 z=\sum^{\infty}_{k=-\infty}\frac{1}{(z-k\pi)^2}\textrm{, }z\neq 0,\pm\pi,\pm 2\pi,\ldots\tag 1 $$ Hence with $z\rightarrow z-\frac{\pi}{2}$, we have $$ \sec^2 z=\sum^{\infty}_{n=-\infty}\frac{1}{(z-\pi/2-k\pi)^2}\tag 2 $$ Also $$ \frac{d}{dt}\int^{\tan t}_{c}\frac{1}{\arctan x}dx=\frac{\sec^2 t}{t} $$ Hence using (1): $$ I=\int^{\tan t}_{c}\frac{1}{\arctan x}dx= $$ $$ =\sum _{k=-\infty }^{\infty } [\frac{4 \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+\frac{8 k \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}- $$ $$ -\frac{4 \pi \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}-\frac{8 k \pi \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+ $$ $$ +\frac{8 t \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+\frac{4 \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+ $$ $$ +\frac{8 k \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}-\frac{8 t \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}]+C. $$ But $$ S_1=\sum^{\infty}_{k=-\infty}\frac{4\pi}{(\pi+2k\pi)^2(\pi+2 k\pi-2t)}=\frac{\pi (-t+\tan t)}{2 t^2} $$ $$ S_2=\sum _{k=-\infty }^{\infty } \frac{8 k \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\pi t-\pi \tan t+2 t \tan t}{2 t^2} $$ $$ S_3=\sum _{k=-\infty }^{\infty } \frac{4 \pi \log (2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\pi \log(2 t) (-t+\tan t)}{2 t^2} $$ $$ S_4=\sum _{k=-\infty }^{\infty } \frac{8 k \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\log(2 t) (\pi t-(\pi -2 t) \tan t)}{2 t^2} $$ $$ S_5=-\sum _{k=-\infty }^{\infty } \frac{8 t \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\log(2 t) (t-\tan t)}{t} $$ Hence $$ I=S_1+S_2+S_3+S_4+S_5-\sum^{\infty}_{k=-\infty}\frac{4\log(\pi+2k\pi-2t)}{\pi^2(2k+1)^2}+C $$ Hence $$ \int^{\tan t}_{c}\frac{dx}{\arctan x}=\log(2t)+\frac{\tan t}{t}-\sum_{k\in\textbf{Z},k-odd}\frac{4\log(\pi k-2t)}{\pi^2 k^2}+C\textrm{, }0<t<\frac{\pi}{2}.\tag 3 $$ The logarithm is defined as $\log x:=\{\log x$, $x>0$ and $i\pi+\log(-x)$, when $x<0\}$.
$$ \int\frac{dx}{\arctan x}=\log\left(2\arctan x\right)+\frac{x}{\arctan x}- $$ $$ -4\sum^{+\infty}_{k=-\infty}\frac{\log\left(\pi (2k+1)-2\arctan x\right)}{\pi^2 (2k+1)^2}+C= $$ $$ =\log\left(2\arctan x\right)+\frac{x}{\arctan x} -4\sum^{+\infty}_{k=-\infty}\frac{\log(\left|\pi (2k+1)-2\arctan x\right|)}{\pi^2 (2k+1)^2}+C,\tag 4 $$ where $x>0$.

Notes.

This is the graph of $\int^{\tan t}_{\tan 1}\frac{1}{\arctan x}dx$ and the evaluation I found where the sum is trancated at $k=5$

At $t=e/2$ cutting the sum in $k=100$, we get

At $t=e/2$ cutting the sum in $k=1000$, we get

This is the graph's of $\int^{t}_{1}\frac{1}{\arctan(x)}dx$ and the evaluations and the evaluation (4):

Long Comment

Similar idea to @YuriNegometyanov:

$$I(x)=\int\limits_1^x \dfrac{\text dt}{\arctan t}=\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{\tan ^2(u)+1}{u} \, du,\tag1$$

$$I(x)=\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{1}{u} \, du+\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{\tan ^2(u)}{u} \, du,\tag2$$

$$I(x)=\log \left(\frac{16 \tan ^{-1}(x)}{\pi^2 }\right)+x^2 \log \left(\tan ^{-1}(x)\right)-2\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \log (u) \tan (u) \sec ^2(u) \, du,\tag3$$

The integral in (3) can also be expressed in terms of the infinite series for $\tan u$, since

$$\frac{d^2 \tan (u)}{d u^2}=2\tan (u) \sec ^2(u)=2 \sum _{k=1}^{\infty } \frac{(2 k-2) (2 k-1) \left(\left(2^{2 k}-1\right) \zeta (2 k)\right) x^{2 k-3}}{\pi ^{2 k}}$$

and

$$\int x^{2 k-3} \log (x) \, dx=\frac{ (2 (k-1) \log (x)-1)}{4 (k-1)^2}x^{2 k-2}$$