If $\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+\mathrm{x}+1\right)^{2}}=\mathrm{a} \tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{b}\left(\frac{2 \mathrm{x}+1}{\mathrm{x}^{2}+\mathrm{x}+1}\right)+\mathrm{C}$, $x>0$ where $C$ is the constant of integration, then the value of $9(\sqrt{3} a+b)$ is equal to
$I=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]^{2}}$
$\int \frac{\mathrm{dt}}{\left(\mathrm{t}^{2}+\frac{3}{4}\right)^{2}}\left(\right.$ Put $\left.\mathrm{x}+\frac{1}{2}=\mathrm{t}\right)$
$=\frac{\sqrt{3}}{2} \int \frac{\sec ^{2} \theta d \theta}{\frac{9}{16} \sec ^{4} \theta}\left(\right.$ Put $\left.t=\frac{\sqrt{3}}{2} \tan \theta\right)$
$=\frac{4 \sqrt{3}}{9} \int(1+\cos 2 \theta) \mathrm{d} \theta$
$=\frac{4 \sqrt{3}}{9}\left[\theta+\frac{\sin 2 \theta}{2}\right]+\mathrm{c}$
$=\frac{4 \sqrt{3}}{9}\left[\tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\frac{\sqrt{3}(2 \mathrm{x}+1)}{3+(2 \mathrm{x}+1)^{2}}\right]+\mathrm{c}$
$=\frac{4 \sqrt{3}}{9} \tan ^{-1}\left(\frac{2 \mathrm{x}+1}{\sqrt{3}}\right)+\frac{1}{3}\left(\frac{2 \mathrm{x}+1}{\mathrm{x}^{2}+\mathrm{x}+1}\right)+\mathrm{c}$
Hence, $9(\sqrt{3} a+b)=15$