Definite Integration

If $\int\limits_{\pi / 6}^{\pi / 4}\left(\cot \left(x-\frac{\pi}{3}\right) \cot \left(x+\frac{\pi}{3}\right)+1\right) d x=\alpha \log _{\mathrm{e}}(\sqrt{3}-1)$, then $9 \alpha^2$ is equal to $\_\_\_\_$ .