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.
Step 1: Understanding the Concept:
To integrate a function of the form $\frac{1}{(ax^2+bx+c)^2}$, we first complete the square of the quadratic in the denominator and then use a trigonometric substitution or the reduction formula for integrals. Step 2: Key Formula or Approach:
Integration of $\frac{1}{(u^2+k^2)^2}$ is $\frac{1}{2k^2} \left[ \frac{u}{u^2+k^2} + \int \frac{du}{u^2+k^2} \right]$. Step 3: Detailed Explanation:
Denominator: $x^2 + x + 1 = (x + 1/2)^2 + 3/4$.
Let $u = x + 1/2 \implies du = dx$. Let $k^2 = 3/4 \implies k = \sqrt{3}/2$.
The integral is $\int \frac{du}{(u^2 + k^2)^2}$.
Using the standard formula:
\[ \int \frac{du}{(u^2 + k^2)^2} = \frac{1}{2k^2} \left( \frac{u}{u^2 + k^2} + \frac{1}{k} \tan^{-1} \frac{u}{k} \right) + C \]
Substitute $k^2 = 3/4$ and $k = \sqrt{3}/2$:
\[ I = \frac{1}{2(3/4)} \left( \frac{x+1/2}{x^2+x+1} + \frac{2}{\sqrt{3}} \tan^{-1} \frac{x+1/2}{\sqrt{3}/2} \right) = \frac{2}{3} \left( \frac{2x+1}{2(x^2+x+1)} + \frac{2}{\sqrt{3}} \tan^{-1} \frac{2x+1}{\sqrt{3}} \right) \]
\[ I = \frac{4}{3\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + \frac{1}{3} \left( \frac{2x+1}{x^2+x+1} \right) + C \]
Comparing with the given form: $a = \frac{4}{3\sqrt{3}} = \frac{4\sqrt{3}}{9}$ and $b = \frac{1}{3}$.
Now, calculate $9(\sqrt{3}a + b)$:
\[ \sqrt{3}a = \sqrt{3} \times \frac{4\sqrt{3}}{9} = \frac{12}{9} = \frac{4}{3} \]
\[ 9(\sqrt{3}a + b) = 9(4/3 + 1/3) = 9(5/3) = 15 \] Step 4: Final Answer:
The value of $9(\sqrt{3}a + b)$ is 15.
