We need to evaluate

$$\\int \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx$$

and compare it with

$$x\sqrt{x^2+x+1}+\alpha\sqrt{x^2+x+1}+\beta\ln\left|x+\frac12+\sqrt{x^2+x+1}\right|+C.$$

We must find $\alpha+2\beta$.

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1. Let

$$Q=x^2+x+1.$$

Then the integral becomes

$$I=\int \frac{2x^2+5x+9}{\sqrt Q}\,dx.$$

We try to rewrite the numerator in terms of $Q$ and $Q' = 2x+1$.

Since

$$2x^2+5x+9 = 2(x^2+x+1) + (3x+7),$$

we get

$$I=\int 2\sqrt Q\,dx + \int \frac{3x+7}{\sqrt Q}\,dx.$$

Now express $3x+7$ as

$$3x+7 = A(2x+1)+B.$$

Comparing coefficients:

$$2A=3 \Rightarrow A=\frac32,$$ $$A+B=7 \Rightarrow \frac32+B=7 \Rightarrow B=\frac{11}{2}.$$

So,

$$3x+7=\frac32(2x+1)+\frac{11}{2}.$$

Hence

$$I=\int 2\sqrt Q\,dx + \frac32\int \frac{2x+1}{\sqrt Q}\,dx + \frac{11}{2}\int \frac{dx}{\sqrt Q}.$$

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2. Evaluate each part

(i) Evaluate $\int \frac{2x+1}{\sqrt Q}\,dx$

Since $dQ=(2x+1)dx$,

$$\int \frac{2x+1}{\sqrt Q}\,dx = \int \frac{dQ}{\sqrt Q}=2\sqrt Q.$$

Therefore,

$$\frac32\int \frac{2x+1}{\sqrt Q}\,dx = 3\sqrt Q.$$

(ii) Evaluate $\int 2\sqrt Q\,dx$

Notice

$$\frac{d}{dx}\big(x\sqrt Q\big)=\sqrt Q + x\cdot \frac{2x+1}{2\sqrt Q} =\frac{2Q+x(2x+1)}{2\sqrt Q}.$$

Since $Q=x^2+x+1$,

$$2Q+x(2x+1)=2(x^2+x+1)+2x^2+x=4x^2+3x+2.$$

Thus,

$$\frac{d}{dx}(x\sqrt Q)=\frac{4x^2+3x+2}{2\sqrt Q}.$$

Then

$$2\sqrt Q = \frac{2Q}{\sqrt Q}=\frac{2x^2+2x+2}{\sqrt Q}.$$

Instead of integrating directly, use the standard reduction:

$$\int 2\sqrt Q\,dx = x\sqrt Q + \frac12\sqrt Q + \frac34\ln\left|x+\frac12+\sqrt Q\right| + C.$$

We will also verify this shortly through the standard formula.

A cleaner way is to complete the square:

$$Q=\left(x+\frac12\right)^2+\frac34.$$

Using the standard result

$$\int \sqrt{u^2+a^2}\,du = \frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln|u+\sqrt{u^2+a^2}|+C,$$

with

$$u=x+\frac12, \qquad a^2=\frac34,$$

we get

$$\int \sqrt Q\,dx = \frac{x+1/2}{2}\sqrt Q + \frac{3}{8}\ln\left|x+\frac12+\sqrt Q\right|+C.$$

Therefore,

$$\int 2\sqrt Q\,dx = \left(x+\frac12\right)\sqrt Q + \frac34\ln\left|x+\frac12+\sqrt Q\right|+C.$$

(iii) Evaluate $\int \frac{dx}{\sqrt Q}$

Again with

$$Q=\left(x+\frac12\right)^2+\frac34,$$

we use the standard formula

$$\int \frac{dx}{\sqrt{x^2+x+1}} = \ln\left|x+\frac12+\sqrt{x^2+x+1}\right|+C.$$

Hence,

$$\frac{11}{2}\int \frac{dx}{\sqrt Q} = \frac{11}{2}\ln\left|x+\frac12+\sqrt Q\right|.$$

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3. Combine all parts

So,

$$I=\left(x+\frac12\right)\sqrt Q + \frac34\ln\left|x+\frac12+\sqrt Q\right| + 3\sqrt Q + \frac{11}{2}\ln\left|x+\frac12+\sqrt Q\right| + C.$$

Combine like terms:

Square-root terms:

$$\left(x+\frac12\right)\sqrt Q + 3\sqrt Q = x\sqrt Q + \frac72\sqrt Q.$$

So,

$$\alpha=\frac72.$$

Log terms:

$$\frac34 + \frac{11}{2} = \frac34 + \frac{22}{4} = \frac{25}{4}.$$

So,

$$\beta=\frac{25}{4}.$$

Thus,

$$\alpha+2\beta = \frac72 + 2\cdot \frac{25}{4} = \frac72 + \frac{25}{2} = \frac{32}{2}=16.$$

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4. Final answer

$$\boxed{16}$$

The derived answer matches the stored correct answer.