Proof $\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}$
Solution 1:
The following formula is a generalization of $(1)$ \begin{align} \small\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{a+ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{ix\sqrt{3}}\right)^2}+e^a\int_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{-ix\sqrt{3}}\right)^2}=1 \end{align} where $|a|$ is sufficiently small. It gives a parametric extension of $(1)$. I think it is hard to expect anything more simple than this.