$\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x=\int \frac{\tan x+\sqrt{3}}{\tan x-\sqrt{3}} d x$
$\begin{aligned} & =\int \frac{\frac{\sin x}{\cos x}+\sqrt{3}}{\frac{\sin x}{\cos x}-\sqrt{3}} d x=\int \frac{\sin x+\sqrt{3} \cos x}{\sin x-\sqrt{3} \cos x} d x \\ & \Rightarrow \quad \sin x+\sqrt{3} \cos x=K_1 \cdot \frac{d}{d x}(\sin x-\sqrt{3} \cos x) \\ & \quad+K_2(\sin x-\sqrt{3} \cos x) \\ & \Rightarrow \quad \sin x+\sqrt{3} \cos x=K_1(\cos x+\sqrt{3} \sin x) \\ & \quad+K_2(\sin x-\sqrt{3} \cos x) \\ & =\left(\sqrt{3} K_1+K_2\right) \sin x+\left(K_1-\sqrt{3} K_2\right) \cos x\end{aligned}$
$\therefore \quad \sqrt{3} K_1+K_2=1$ ...(i)
$K_1-\sqrt{3} K_2=\sqrt{3}$ ...(ii)
$\therefore \quad K_1=\sqrt{3}+\sqrt{3} K_2$
From (i), $\sqrt{3}\left(\sqrt{3}+\sqrt{3} K_2\right)+K_2=1$
$\begin{aligned} & \Rightarrow \quad 3+4 K_2=1 \Rightarrow K_2=\frac{-1}{2} \\ & \therefore \quad K_1=\sqrt{3}\left(1+K_2\right)=\frac{\sqrt{3}}{2} \\ & \therefore \quad \int \frac{\frac{\sqrt{3}}{2}(\cos x+\sqrt{3} \sin x)-\frac{1}{2}(\sin x-\sqrt{3} \cos x)}{(\sin x-\sqrt{3} \cos x)} d x \\ & =\frac{\sqrt{3}}{2} \int \frac{\cos x+\sqrt{3} \sin x}{\sin x-\sqrt{3} \cos x} d x-\frac{1}{2} \int \frac{\sin x-\sqrt{3} \cos x}{\sin x-\sqrt{3} \cos x} d x \\ & =\frac{-1}{2} x+\frac{\sqrt{3}}{2} \ln |\sin x-\sqrt{3} \cos x|+C \\ & =\frac{-1}{2} x+\frac{\sqrt{3}}{2} \ln 2\left|\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x\right|+C \\ & =\frac{-1}{2} x+\frac{\sqrt{3}}{2} \ln 2\left|\sin \left(x-\frac{\pi}{3}\right)\right|+C \\ & =\frac{-1}{2} x+\frac{\sqrt{3}}{2} \ln \left|\sin \left(x-\frac{\pi}{3}\right)\right|+\frac{\sqrt{3}}{2} \ln 2+C \\ & =\frac{-1}{2} x+\frac{\sqrt{3}}{2} \ln \left|\sin \left(x-\frac{\pi}{3}\right)\right|+C\end{aligned}$