Remarkable integral: $\int_0^{\infty} x \left(1 - \frac{\sinh x}{\cosh x-\sqrt 3/2} \right) = -\frac{13 \pi ^2}{72}$?

Integration by parts does help: as $$1-\frac{\sinh x}{\cosh x-\cos\gamma}=\left(-\ln\frac{\cosh x-\cos\gamma}{e^x/2}\right)',$$ after IbP and further change of variables $t=e^{-x}$ the integral transforms into $$\int_0^{1}\frac{\ln\left(1-e^{i\gamma}t\right)\left(1-e^{-i\gamma}t\right)}{t}dt=-\left[2\Re\operatorname{Li}_2\left(e^{i\gamma}t\right)\right]_{0}^{1}=-2\Re\operatorname{Li}_2\left(e^{i\gamma}\right),$$ with $\gamma=\frac{\pi}{6}$. Now to get the result it remains to use the formula $$\Re\,\mathrm{Li}_2\left(e^{i\gamma}\right)=\frac{\gamma^2}{4}-\frac{\pi\gamma}{2}+\frac{\pi^2}6,\qquad \gamma\in(0,2\pi).$$ This also yields the conjecture mentioned in the comments.

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Added: We can also obtain the result pretending that we don't know anything about dilogarithms. Namely, differentiate the integral with respect to parameter $\gamma$: \begin{align} \frac{\partial}{\partial \gamma}\int_0^{1}\frac{\ln\left(1-e^{i\gamma}t\right)\left(1-e^{-i\gamma}t\right)}{t}dt&=-i\int_0^1\left[\frac{e^{i\gamma}}{1-e^{i\gamma}t}-\frac{e^{-i\gamma}}{1-e^{-i\gamma}t}\right]dt=\\&=i\biggl[\ln\frac{1-e^{i\gamma}t}{1-e^{-i\gamma}t}\biggr]_0^1=-2 \biggl[\operatorname{arg}(1-e^{i\gamma}t)\biggr]_0^1=\\&=-2\biggl[\left(\frac{\gamma}{2}-\frac{\pi}{2}\right)-0\biggr]=\pi-\gamma, \end{align} where we again assume that $\gamma\in(0,2\pi)$. We can now integrate back with respect to $\gamma$ to get the previously obtained formula using that for $\gamma=0$ our integral reduces to computation of $\zeta(2)$ (expand the integrand into Taylor series w.r.t. $t$).

The following is another approach that also involves differentiating under the integral sign.

Let $$I(\theta) = \int_{0}^{\infty} x \left(1- \frac{\sinh x}{\cosh x - \cos \theta} \right) \, \mathrm dx \, , \quad 0 < \theta < \pi $$

Then $$I'(\theta) = \sin \theta \int_{0}^{\infty} \frac{x \sinh x}{\left(\cosh x - \cos \theta \right)^{2}} \, \mathrm dx$$

I tried evaluating $I'(\theta)$ using contour integration, but calculating the residues became too tedious.

Instead let $a$ be positive parameter and define $$J(\alpha) = \int_{0}^{\infty}\frac{\mathrm d x}{\cosh (\alpha x)- \cos \theta} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{du}{\cosh u - \cos \theta}$$

$I'(\theta)$ is then $ -\sin (\theta) \, J'(1) $.

To evaluate $J(\alpha)$, we can integrate the function $f(z) = \frac{z}{\cosh z - \cos \theta}$ around a rectangle contour in the upper-half complex plane of height $2 \pi i $.

We end up with $$- 2 \pi i \int_{-\infty}^{\infty} \frac{dt}{\cosh t - \cos \theta} = 2 \pi i \, \left(\operatorname{Res} \left[f(z), i \theta \right]+ \operatorname{Res} \left[f(z), i \left(2 \pi - \theta\right)\right] \right)= 4 \pi i \, \frac{\left(\theta - \pi\right)}{\sin \theta}$$

Therefore, $J(\alpha) = \frac{1}{\alpha}\frac{\pi - \theta}{\sin \theta} $, which means $I'(\theta) =\pi - \theta $.

Integrating with respect to $\theta$, we get $$I(\theta) = \pi \theta -\frac{\theta^{2}}{2} + C $$

To determine the constant of integration, we can evaluate $I(\theta)$ at $\theta = \frac{\pi}{2}$.

$$\begin{align} I(\pi/2) &= \int_{0}^{\infty} x \left(1- \tanh x \right) \mathrm dx \\ &= (1- \tanh x) \frac{x^{2}}{2}\Bigg|^{\infty}_{0} + \frac{1}{2}\int_{0}^{\infty}\frac{x^{2}}{\cosh^{2} x} \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{x^{2}}{\cosh^{2} x} \mathrm dx \\ &= 2 \int_{0}^{\infty}x^{2} \, \frac{e^{-2x}}{(1+e^{-2x})^{2}} \, \mathrm dx \\ &= 2 \int_{0}^{\infty}x^{2} \, \sum_{n=1}^{\infty} (-1)^{n-1} n e^{-2nx} \mathrm dx \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} n \int_{0}^{\infty} x^{2} e^{-2nx} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{2}} = \frac{\pi^{2}}{24} \end{align}$$

So $C= \frac{\pi^{2}}{24} -\frac{\pi^2}{2} + \frac{\pi^{2}}{8} = -\frac{\pi^2}{3}$, and, therefore, $$I(\theta) = \pi \theta -\frac{\theta^{2}}{2} - \frac{\pi^{2}}{3} $$