If $\int \frac{d x}{\left(x^2+x+1\right)^2}=a \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+b\left(\frac{2 x+1}{x^2+x+1}\right)+C, x>0$ where $C$ is the constant of integration, then the value of $9(\sqrt{3} a+b)$ is equal to $\_\_\_\_$。
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