If ∫ 2 x^2+5 x+9 √x^2+x+1 ~d x=x √x^2+x+1+α √x^2+x+1+β log _e |x+ 1 2+√x^2+x+1 |+ C, where C is the constant of integration, then α+2 β is equal to .

$$\begin{aligned} & 2 \mathrm{x}^2+5 \mathrm{x}+9=\mathrm{A}\left(\mathrm{x}^2+\mathrm{x}+1\right)+\mathrm{B}(2 \mathrm{x}+1)+\mathrm{C} \\ & \mathrm{A}=2 \quad \mathrm{~B}=\frac{3}{2} \quad \mathrm{C}=\frac{11}{2} \\ & 2 \int \sqrt{\mathrm{x}^2+\mathrm{x}+1} \mathrm{dx}+\frac{3}{2} \int \frac{2 \mathrm{x}+1}{\sqrt{\mathrm{x}^2+\mathrm{x}+1}} \mathrm{dx}+\frac{11}{2} \int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2+\mathrm{x}+1}} \\ & 2 \int \sqrt{\left(\mathrm{x}+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \mathrm{dx}+3 \sqrt{\mathrm{x}^2+\mathrm{x}+1}+\frac{11}{2} \int \frac{\mathrm{dx}}{\sqrt{\left(\mathrm{x}+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}} \\ & 2\left(\frac{\mathrm{x}+\frac{1}{2}}{2} \sqrt{\mathrm{x}^2+\mathrm{x}+1}+\frac{3}{8} \ln \left(\mathrm{x}+\frac{1}{2}+\sqrt{\mathrm{x}^2+\mathrm{x}+1}\right)\right)+3 \sqrt{\mathrm{x}^2+\mathrm{x}+1} \\ & +\frac{11}{2} \ell \mathrm{n}\left(\mathrm{x}+\frac{1}{2}+\sqrt{\mathrm{x}^2+\mathrm{x}+1}\right)+\mathrm{C} \\ & \alpha=\frac{7}{2} \quad\beta=\frac{25}{4} \\& \alpha+2 \beta=16 \\ & \end{aligned}$$​2x2+5x+9=A(x2+x+1)+B(2x+1)+CA=2 B=23​C=211​2∫x2+x+1​dx+23​∫x2+x+1​2x+1​dx+211​∫x2+x+1​dx​2∫(x+21​)2+(23​​)2​dx+3x2+x+1​+211​∫(x+21​)2+(23​​)2​dx​2(2x+21​​x2+x+1​+83​ln(x+21​+x2+x+1​))+3x2+x+1​+211​ℓn(x+21​+x2+x+1​)+Cα=27​β=425​α+2β=16​