If int frac{d x}{left(x^{2}+x+1right)^{2}}=a | vedclass

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(B) Let $I = \int \frac{dx}{(x^2+x+1)^2} = \int \frac{dx}{((x+1/2)^2 + 3/4)^2}$.Substitute $t = x + 1/2$,so $dt = dx$. Then $I = \int \frac{dt}{(t^2 +

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$15$

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(B) Let $I = \int \frac{dx}{(x^2+x+1)^2} = \int \frac{dx}{((x+1/2)^2 + 3/4)^2}$.Substitute $t = x + 1/2$,so $dt = dx$. Then $I = \int \frac{dt}{(t^2 + 3/4)^2}$.Let $t = \frac{\sqrt{3}}{2} 	an 	heta$,then $dt = \frac{\sqrt{3}}{2} \sec^2 	heta d	heta$.$I = \int \frac{\frac{\sqrt{3}}{2} \sec^2 	heta d	heta}{(\frac{3}{4} 	an^2 	heta + 3/4)^2} = \int \frac{\frac{\sqrt{3}}{2} \sec^2 	heta d	heta}{\frac{9}{16} \sec^4 	heta} = \frac{8\sqrt{3}}{9} \int \cos^2 	heta d	heta$.Using $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get $I = \frac{4\sqrt{3}}{9} \int (1 + \cos 2	heta) d	heta = \frac{4\sqrt{3}}{9} (	heta + \frac{\sin 2	heta}{2}) + C$.Since $	an 	heta = \frac{2t}{\sqrt{3}} = \frac{2x+1}{\sqrt{3}}$,we have $	heta = 	an^{-1}(\frac{2x+1}{\sqrt{3}})$.Also,$\frac{\sin 2	heta}{2} = \frac{	an 	heta}{1 + 	an^2 	heta} = \frac{\frac{2x+1}{\sqrt{3}}}{1 + \frac{(2x+1)^2}{3}} = \frac{\sqrt{3}(2x+1)}{3 + 4x^2 + 4x + 1} = \frac{\sqrt{3}(2x+1)}{4(x^2+x+1)}$.Thus,$I = \frac{4\sqrt{3}}{9} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + \frac{4\sqrt{3}}{9} \cdot \frac{\sqrt{3}(2x+1)}{4(x^2+x+1)} + C = \frac{4\sqrt{3}}{9} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + \frac{1}{3} \frac{2x+1}{x^2+x+1} + C$.Comparing with the given form,$a = \frac{4\sqrt{3}}{9}$ and $b = \frac{1}{3}$.Then $9(\sqrt{3}a + b) = 9(\sqrt{3} \cdot \frac{4\sqrt{3}}{9} + \frac{1}{3}) = 9(\frac{12}{9} + \frac{1}{3}) = 9(\frac{4}{3} + \frac{1}{3}) = 9(\frac{5}{3}) = 15$.

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

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