= Solution
The <Cauchy principal value> deletes symmetric intervals around the real poles $-1$ and $1$ and takes a symmetric <limit> at infinity:
$$
\operatorname{PV}\int_{-\infty}^{\infty}f(x)dx
=\lim_{R\to\infty,\ \varepsilon\downarrow0}
\left(\int_{-R}^{-1-\varepsilon}+\int_{-1+\varepsilon}^{1-\varepsilon}+\int_{1+\varepsilon}^{R}\right)f(x)dx.
$$
The integrand is even. The standard <principal-value beta integral> gives
$$
\operatorname{PV}\int_0^\infty\frac{x^{s-1}}{1-x^a}dx=\frac\pi a\cot\frac{\pi s}{a}.
$$
Taking $s=1$, $a=6$, and reversing the denominator,
$$
\operatorname{PV}\int_{-\infty}^{\infty}\frac{dx}{x^6-1}
=-\frac{2\pi}{6}\cot\frac\pi6=-\frac\pi{\sqrt3}.
$$
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