If $\displaystyle\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)+1\right)dx = \alpha\log_e(\sqrt{3}-1)$, then $9\alpha^2$ is equal to ________.

Given integral is $I = \displaystyle\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)+1\right)dx$

Using the identity $\cot A \cot B + 1 = \dfrac{\cos(A-B)}{\sin A \sin B}$, the integrand simplifies to:

$f(x) = \dfrac{\cos\left(\left(x-\dfrac{\pi}{3}\right) - \left(x+\dfrac{\pi}{3}\right)\right)}{\sin\left(x-\dfrac{\pi}{3}\right)\sin\left(x+\dfrac{\pi}{3}\right)}$

$f(x) = \dfrac{\cos\left(-\dfrac{2\pi}{3}\right)}{\sin^2 x - \sin^2\left(\dfrac{\pi}{3}\right)}$

$f(x) = \dfrac{-1/2}{\sin^2 x - 3/4} = \dfrac{2}{3 - 4\sin^2 x}$

Dividing the numerator and the denominator by $\cos^2 x$:

$f(x) = \dfrac{2\sec^2 x}{3\sec^2 x - 4\tan^2 x} = \dfrac{2\sec^2 x}{3(1+\tan^2 x) - 4\tan^2 x} = \dfrac{2\sec^2 x}{3 - \tan^2 x}$

The integral becomes:

$I = \displaystyle\int_{\pi/6}^{\pi/4} \dfrac{2\sec^2 x}{3 - \tan^2 x} dx$

Substituting $\tan x = t \Rightarrow \sec^2 x dx = dt$.

When $x = \dfrac{\pi}{6}$, $t = \dfrac{1}{\sqrt{3}}$. When $x = \dfrac{\pi}{4}$, $t = 1$.

$I = \displaystyle\int_{1/\sqrt{3}}^{1} \dfrac{2}{3 - t^2} dt$

Using the standard integral $\displaystyle\int \dfrac{1}{a^2 - x^2} dx = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right|$:

$I = 2 \left[ \dfrac{1}{2\sqrt{3}} \ln\left|\dfrac{\sqrt{3}+t}{\sqrt{3}-t}\right| \right]_{1/\sqrt{3}}^{1}$

$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln\left(\dfrac{\sqrt{3}+\dfrac{1}{\sqrt{3}}}{\sqrt{3}-\dfrac{1}{\sqrt{3}}}\right) \right)$

$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln\left(\dfrac{3+1}{3-1}\right) \right)$

$I = \dfrac{1}{\sqrt{3}} \left( \ln\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \ln 2 \right) = \dfrac{1}{\sqrt{3}} \ln\left(\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)}\right)$

Rationalizing the term inside the logarithm by multiplying the numerator and the denominator by $(\sqrt{3}-1)$:

$\dfrac{\sqrt{3}+1}{2(\sqrt{3}-1)} = \dfrac{(\sqrt{3}+1)(\sqrt{3}-1)}{2(\sqrt{3}-1)^2} = \dfrac{3-1}{2(\sqrt{3}-1)^2} = \dfrac{1}{(\sqrt{3}-1)^2} = (\sqrt{3}-1)^{-2}$

Thus,

$I = \dfrac{1}{\sqrt{3}} \ln\left((\sqrt{3}-1)^{-2}\right) = -\dfrac{2}{\sqrt{3}} \ln(\sqrt{3}-1)$

Comparing this with $I = \alpha\log_e(\sqrt{3}-1)$, we get:

$\alpha = -\dfrac{2}{\sqrt{3}}$

Therefore, $9\alpha^2 = 9 \left(-\dfrac{2}{\sqrt{3}}\right)^2 = 9 \times \dfrac{4}{3} = 12$.

Answer: $12$