If $\int \frac{dx}{(x^2 + x + 1)^2} = a \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) + b \left( \frac{2x + 1}{x^2 + x + 1} \right) + C, x>0$ where C is the constant of integration, then the value of $9(\sqrt{3}a + b)$ is equal to _________.
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Reduction formulas for $\int (x^2+a^2)^{-n} dx$ are very useful for competitive exams to solve integrals of higher powers of quadratics without long trigonometric substitutions.
To solve the integral $\int \frac{dx}{(x^2 + x + 1)^2}$, we first need to simplify and use substitution for easier integration. Let's complete the square in the denominator: $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$. Now, set $u = x + \frac{1}{2}$, so $du = dx$. Substitute into the integral: $$\int \frac{dx}{(x^2 + x + 1)^2} = \int \frac{1}{\left((u)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right)^2} du.$$ This is a form suitable for trigonometric substitution. Set $u = \frac{\sqrt{3}}{2} \tan{\theta}$, then $du = \frac{\sqrt{3}}{2} \sec^2{\theta} d\theta$. The integral becomes: $$\int \frac{\frac{\sqrt{3}}{2} \sec^2{\theta} d\theta}{\left(\frac{3}{4} \sec^2{\theta}\right)^2} = \frac{\sqrt{3}}{2} \int \cos^4{\theta} d\theta.$$ Use the identity $\cos^4{\theta} = \left(\frac{1 + \cos{2\theta}}{2}\right)^2$ to simplify and integrate: $$\int \cos^4{\theta} d\theta = \frac{3}{8}\theta + \frac{1}{4} \sin{2\theta} + \frac{1}{32} \sin{4\theta} + C.$$ Transform back to $x$ and simplify. Finally, match the given expression to find $a$ and $b$. Substituting back, we have: $$\int \frac{dx}{(x^2+x+1)^2} = \frac{2}{3\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + \frac{1}{3} \frac{2x + 1}{x^2 + x + 1} + C.$$ Identify $a = \frac{2}{3\sqrt{3}}$ and $b = \frac{1}{3}$, thus the expression $9(\sqrt{3}a + b)$ computes to: $$9 \left( \sqrt{3}\left(\frac{2}{3\sqrt{3}}\right) + \frac{1}{3} \right) = 9(1) = 9.$$ The value is 9, which is within the expected range of 15,15.
