A conjecture $\int_{-\infty}^\infty\frac{\arctan e^x}{\cosh x}\cdot\frac{\tanh\frac{x}2}{x}dx\stackrel?=\frac\pi2\ln2$ I need to find a closed form for this integral:
$$\mathcal{I}=\int_{-\infty}^\infty\frac{\arctan e^x}{\cosh x}\cdot\frac{\tanh\frac{x}2}{x}dx.$$
A numerical integration results in an approximation $\mathcal{I}\approx1.0887930451518...$, and WolframAlpha suggests a possible closed form for this number:
$$\mathcal{I}\stackrel?=\frac\pi2\ln2$$

Is it the correct value of this integral? If so, how to prove it?

 A: Replacing $x \mapsto -x$, we have
$$ I = \int_{-\infty}^{\infty} \frac{\frac{\pi}{2} - \arctan e^{x}}{\cosh x} \, \tanh\left(\frac{x}{2}\right) \, \frac{dx}{x}. $$
Averaging,
\begin{align*}
I
&= \frac{\pi}{4} \int_{-\infty}^{\infty} \frac{\tanh(x/2)}{\cosh x} \, \frac{dx}{x} \\
&= \pi \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-x})}{(1 + e^{-2x})(1 + e^{-x})} \, \frac{dx}{x} \\
&= \pi \int_{0}^{\infty} \left( \frac{e^{-x}}{1+ e^{-x}} - \frac{e^{-2x}}{1 + e^{-2x}} \right) \, \frac{dx}{x}. \tag{1}
\end{align*}
Note that the last integral belongs to the class of integrals called Frullani integral:
$$ \int_{0}^{\infty} \frac{f(bx) - f(ax)}{x} \, dx. \tag{2} $$
Heuristically, we can evaluate (2) as
\begin{align*}
\int_{0}^{\infty} \frac{f(bx) - f(ax)}{x} \, dx
&= \int_{0}^{\infty} \int_{a}^{b} f'(xt) \, dtdx
 = \int_{a}^{b} \int_{0}^{\infty} f'(xt) \, dxdt \\
&= \int_{a}^{b} \frac{f(\infty) - f(0)}{t} dt
 = \{ f(\infty) - f(0) \} \log\left(\frac{b}{a}\right)
\end{align*}
and this is justified via Fubini's theorem if $f$ is absoultely continuous and $f'$ is either integrable or non-negative on $[0, \infty)$, which is clearly the case for 
$$f(x) = \frac{e^{-x}}{1 + e^{-x}}$$
with $(a, b) = (1, 2)$. Thus it follows that
\begin{align*}
\int_{0}^{\infty} \left( \frac{e^{-x}}{1+ e^{-x}} - \frac{e^{-2x}}{1 + e^{-2x}} \right) \, \frac{dx}{x}
&= \frac{1}{2}\log 2.
\end{align*}
Plugging this back to (1), we obtain the desired result.
