$$\text{If }\int \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx=x\sqrt{x^2+x+1}+\alpha\sqrt{x^2+x+1}+\beta\log_e\!\left|x+\frac12+\sqrt{x^2+x+1}\right|+C,\text{where } C \text{ is the constant of integration,}\text{ then } \alpha+2\beta \text{ is equal to }\underline{\hspace{2cm}}.$$

We need: $$\int \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx = x\sqrt{x^2+x+1} + \alpha\sqrt{x^2+x+1} + \beta\log_e\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| + C$$

Differentiate the RHS:

$$\frac{d}{dx}\left[x\sqrt{x^2+x+1}\right] = \sqrt{x^2+x+1} + x \cdot \frac{2x+1}{2\sqrt{x^2+x+1}}$$

$$= \frac{x^2+x+1+x(2x+1)/2}{\sqrt{x^2+x+1}} = \frac{2(x^2+x+1)+x(2x+1)}{2\sqrt{x^2+x+1}} = \frac{4x^2+3x+2}{2\sqrt{x^2+x+1}}$$

$$\frac{d}{dx}\left[\alpha\sqrt{x^2+x+1}\right] = \alpha \cdot \frac{2x+1}{2\sqrt{x^2+x+1}}$$

$$\frac{d}{dx}\left[\beta\log_e\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|\right] = \frac{\beta}{\sqrt{x^2+x+1}}$$

Adding: $$\frac{4x^2+3x+2 + \alpha(2x+1) + 2\beta}{2\sqrt{x^2+x+1}} = \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}$$

So: $$4x^2+3x+2+2\alpha x+\alpha+2\beta = 2(2x^2+5x+9) = 4x^2+10x+18$$

Comparing coefficients:

$$x^2$$: $$4 = 4$$ (ok)

$$x$$: $$3 + 2\alpha = 10$$, so $$\alpha = 7/2$$

constant: $$2 + \alpha + 2\beta = 18$$, so $$2\beta = 18 - 2 - 7/2 = 16 - 7/2 = 25/2$$, giving $$\beta = 25/4$$

$$\alpha + 2\beta = 7/2 + 25/2 = 32/2 = 16$$.

The answer is 16.