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

$\int \frac{dx}{(x^2+x+1)}=a{\tan }^{-1}\left(\frac{{\displaystyle 2x+1}}{{\displaystyle \sqrt{3}}}\right)+b\left(\frac{{\displaystyle 2x+1}}{{\displaystyle x^2+x+1}}\right)$

$\frac{1}{{(x^2+x+1)}^2}=a\frac{1}{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}\times \frac{2}{\sqrt{3}}+b\left(\frac{{\displaystyle (x^2+x+1)2-(2x+1)(2x+1)}}{{\displaystyle {(x^2+x+1)}^2}}\right)$

$\frac{1}{{(x^2+x+1)}^2}=a.\frac{2}{\sqrt{3}}.\frac{1}{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}+b\left(\frac{{\displaystyle -2x^2-2x+1}}{{\displaystyle {(x^2+x+1)}^2}}\right)$

Put x=0

$1=a.\frac{2}{\sqrt{3}}.\frac{3}{4}+b\left(1\right)$

$\Rightarrow \frac{\sqrt{3}}{2}a+b=1\_\_\_\_\left(1\right)\Rightarrow b=1-\frac{\sqrt{3}}{2}a$

Put x=1

$\frac{1}{9}=\frac{2}{\sqrt{3}}.\frac{a}{4}+b\left(\frac{{\displaystyle -3}}{{\displaystyle 9}}\right)\_\_\_\left(2\right)$

$\frac{1}{9}=\frac{a}{2\sqrt{3}}+\left(-1+\frac{{\displaystyle \sqrt{3}a}}{{\displaystyle 2}}\right)\frac{1}{3}$

$\frac{1}{9}=\frac{a}{2\sqrt{3}}+\frac{a}{2\sqrt{3}}-\frac{1}{3}$

$\Rightarrow \frac{1}{9}+\frac{1}{3}=\frac{2a}{2\sqrt{3}}$

$\Rightarrow \frac{4}{9}=\frac{a}{\sqrt{3}}\Rightarrow a=\frac{4}{3\sqrt{3}}$

$b=1-\frac{\sqrt{3}}{2}a$

$b=1-\frac{\sqrt{3}}{2}\left(\frac{{\displaystyle 4}}{{\displaystyle 3\sqrt{3}}}\right)$

$\Rightarrow b=1-\frac{2}{3}$

$b=\frac{1}{3}$

$9\left(\sqrt{3}a+b\right)=9\left(\sqrt{3}.\frac{4}{3\sqrt{3}}+\frac{1}{3}\right)$

$\Rightarrow 9\left(\frac{4}{3}+\frac{1}{3}\right)$

$\Rightarrow 9\left(\sqrt{3}a+b\right)=15$