If $\displaystyle\int \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx = x\sqrt{x^2+x+1} + \alpha\sqrt{x^2+x+1} + \beta\log_e\!\left|x+\tfrac{1}{2}+\sqrt{x^2+x+1}\right| + C$, where $C$ is the constant of integration, then $\alpha + 2\beta$ is equal to ____.
Step-by-Step Solution
Key Concept: Write $2x^2+5x+9 = A(x^2+x+1) + B(2x+1) + C$ to match with $\sqrt{x^2+x+1}$, $\tfrac{d}{dx}(x^2+x+1)$, and the standard $\int 1/\sqrt{\cdot}\,dx$ form, then extract $A$, $B$, $C$ and read off $\alpha$, $\beta$.
Write $2x^2+5x+9 = A(x^2+x+1)+B(2x+1)+C_0$.
Comparing: $A = 2$, $B = \tfrac{3}{2}$, $C_0 = \tfrac{11}{2}$.
$$\int\!\frac{2x^2+5x+9}{\sqrt{x^2+x+1}}dx = 2\int\!\sqrt{x^2+x+1}\,dx + \frac{3}{2}\int\!\frac{2x+1}{\sqrt{x^2+x+1}}dx + \frac{11}{2}\int\!\frac{dx}{\sqrt{x^2+x+1}}.$$
Using standard results and simplifying:
$$\alpha = \frac{7}{2},\quad \beta = \frac{25}{4}.$$
$$\alpha + 2\beta = \frac{7}{2} + \frac{25}{2} = 16.$$
Correct Answer: 16