If ∫ dx (x^2+x+1)^2=a tan^-1( 2x+1 √3)+b( 2x+1 x^2+x+1)+C,x>0 where C is the constant of integration, then the value of 9(√3a+b) is equal to .

$$I=\int \frac{dx}{{[{(x+\frac{1}{2})}^{2}+\frac{3}{4}]}^{2}}$$I=∫[(x+21​)2+43​]2dx​

Put $x+\frac{1}{2}=t$x+21​=t

$$=\int \frac{dt}{{({t}^{2}+\frac{3}{4})}^{2}}$$=∫(t2+43​)2dt​

$$[Putt=\frac{\sqrt{3}}{2}\mathrm{tan}\theta ] =\frac{\sqrt{3}}{2}\int \frac{{\mathrm{sec}}^{2}\theta d\theta }{\frac{9}{16}{\mathrm{sec}}^{4}\theta }$$[Putt=23​​tanθ]=23​​∫169​sec4θsec2θdθ​

$$=\frac{4\sqrt{3}}{9}\int (1+\mathrm{cos}2\theta )d\theta$$=943​​∫(1+cos2θ)dθ

$$=\frac{4\sqrt{3}}{9}[\theta +\frac{\mathrm{sin}2\theta }{2}]+C$$=943​​[θ+2sin2θ​]+C

$$=\frac{4\sqrt{3}}{9}[{\mathrm{tan}}^{-1}(\frac{2x+1}{3})+\frac{\sqrt{3}(2x+1)}{3+{(2x+1)}^{2}}]+C$$=943​​[tan−1(32x+1​)+3+(2x+1)23​(2x+1)​]+C

$$=\frac{4\sqrt{3}}{9}{\mathrm{tan}}^{-1}(\frac{2x+1}{\sqrt{3}})+\frac{1}{3}(\frac{2x+1}{{x}^{2}+x+1})+C$$=943​​tan−1(3​2x+1​)+31​(x2+x+12x+1​)+C

Hence, $9(\sqrt{3}a+b)=15$9(3​a+b)=15