= Solution
Take $g(z)=L(z)^3/z^3$ on the same <logarithm branch>. At zero, $L(z)^3=-z^3+O(z^4)$, so the apparent singularity is removable and the closed <contour integral> is zero. The bank difference is
$$
(\ell-i\pi)^3-(\ell+i\pi)^3=-6i\pi\ell^2+2i\pi^3.
$$
The circular contributions vanish by the previous estimates with the third power of the logarithm. Consequently
$$
0=-6i\pi\int_1^\infty\frac{\log^2(x-1)}{x^3}\,dx+2i\pi^3\int_1^\infty\frac{dx}{x^3}.
$$
The last <integral> is $1/2$, giving
$$
\boxed{\int_1^\infty\frac{\log^2(x-1)}{x^3}\,dx=\frac{\pi^2}{6}.}
$$
Increasing the logarithm power has converted the desired real <integral> into a jump across a <branch cut>, while removing the pole inside the contour.
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