$$\newcommand{\arctanh}{~\mathrm{arctanh}~}\newcommand{\sech}{~\mathrm{sech}~}$$
$$I=\int_{-1}^1\frac{\arctan x}{\arctanh x}\,\mathrm{d}x$$
Mathematica gives an approximate result of $I=1.581949621806183890451628...$, but no exact form. I predict it's a function of $e$ and $\pi$, and perhaps even the Golden Ratio $\phi$ (It certainly wouldn't be the first time)
The motivation behind this question is pure curiosity. I thought the shape looked nice :)
1st edit: Substitutions of $x=\tan u$ and $x=\tanh u$ respectively yield
$$I= 2\int_{-\pi/4}^{\pi/4}\dfrac{u\sec^2u}{\ln|\frac{1+\tan u}{1-\tan u}|}\,\mathrm{d}u$$
$$I= \int_{-\infty}^{\infty}\dfrac{\arctan(\tanh u)}{u}\sech^2u\,\mathrm{d}u$$
2nd edit: I've considered another approach starting with parameterizing the desired integral by
$$I_a=\int_{-1}^1\frac{\arctan ax}{\arctanh x}\,\mathrm{d}x$$
so that
$$\frac{\partial I_a}{\partial a}=\int_{-1}^1\frac{x}{(1+(ax)^2)\arctanh x}\,\mathrm{d}x$$
Integrating by parts with
$$\begin{matrix}u=\dfrac{1}{\arctanh x}&&\mathrm{d}v=\dfrac{x}{1+(ax)^2}\,\mathrm{d}x\\[1ex]
\mathrm{d}u=\dfrac{\mathrm{d}x}{(x^2-1)\arctanh^2x}&&v=\dfrac{1}{2a^2}\log(1+(ax)^2)\end{matrix}$$
yields the following integral:
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-1}^1\frac{\log(1+(ax)^2)}{(1-x^2)\arctanh^2x}\,\mathrm{d}x$$
which can be modified by a substitution of $y=\arctanh x$ to obtain
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-\infty}^\infty \frac{\log(1+(a\tanh y)^2)}{y^2}\,\mathrm{d}y$$
I have an idea of approaching the remaining integral using the series expansion of $\log(1+x)$; namely, the integral would become
$$\frac{\partial I_a}{\partial a}=\frac{1}{2a^2}\int_{-\infty}^\infty \frac{\mathrm{d}y}{y^2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(a\tanh y)^{2k}=-\frac{1}{2a^2}\sum_{k=1}^\infty \frac{a^{2k}(-1)^k}{k}\underbrace{\int_{-\infty}^\infty \frac{\tanh^{2k}y}{y^2}\,\mathrm{d}y}_{J_k}$$
According to this question, we have a closed from $J_k$ in the case of $k=1$ and potentially all $k>1$ in terms of the Riemann zeta function, but I have yet to do any more investigation.
Another method that occurred to me was to consider a keyhole contour to tackle $\dfrac{\partial I_a}{\partial a}$ but I'm afraid I'm not familiar enough with complex analysis to make that jump just yet.
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