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

$I=\int \frac{dx}{{\left[{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle \frac{3}{4}}\right]}^2}$

Put $x+\frac{1}{2}=t$

$=\int \frac{dt}{{\left(t^2+{\displaystyle \frac{3}{4}}\right)}^2}$

$$\begin{gathered} \left[Put \mathrm{t}=\frac{\sqrt{3}}{2}\tan \theta \right] \\ =\frac{\sqrt{3}}{2}\int \frac{{\sec }^2\theta d\theta }{{\displaystyle \frac{9}{16}}{\sec }^4\theta } \end{gathered}$$

$=\frac{4\sqrt{3}}{9}\int \left(1+\cos 2\theta \right)d\theta$

$=\frac{4\sqrt{3}}{9}\left[\theta +\frac{\sin 2\theta }{2}\right]+C$

$=\frac{4\sqrt{3}}{9}\left[{\tan }^{-1}\left(\frac{2x+1}{3}\right)+\frac{\sqrt{3}\left(2x+1\right)}{3+{\left(2x+1\right)}^2}\right]+C$

$=\frac{4\sqrt{3}}{9}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{1}{3}\left(\frac{2x+1}{x^2+x+1}\right)+C$

Hence, $9\left(\sqrt{3}a+b\right)=15$