JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 22)

If $\int \frac{2 x^2+5 x+9}{\sqrt{x^2+x+1}} \mathrm{~d} x=x \sqrt{x^2+x+1}+\alpha \sqrt{x^2+x+1}+\beta \log _{\mathrm{e}}\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+\mathrm{C}$, where $C$ is the constant of integration, then $\alpha+2 \beta$ is equal to __________ .
Answer
16

Explanation

$$\begin{aligned} & 2 \mathrm{x}^2+5 \mathrm{x}+9=\mathrm{A}\left(\mathrm{x}^2+\mathrm{x}+1\right)+\mathrm{B}(2 \mathrm{x}+1)+\mathrm{C} \\ & \mathrm{~A}=2 \quad \mathrm{~B}=\frac{3}{2} \quad \mathrm{C}=\frac{11}{2} \end{aligned}$$

$$\begin{aligned} & 2 \int \sqrt{x^2+x+1} d x+\frac{3}{2} \int \frac{2 x+1}{\sqrt{x^2+x+1}} d x+\frac{11}{2} \int \frac{d x}{\sqrt{x^2+x+1}} \\ & 2 \int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \mathrm{dx}+3 \sqrt{x^2+x+1}+\frac{11}{2} \int \frac{\mathrm{dx}}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}} \\ & 2\left(\frac{x+\frac{1}{2}}{2} \sqrt{x^2+x+1}+\frac{3}{8} \ell n\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right)\right)+3 \sqrt{x^2+x+1} \\ & +\frac{11}{2} \ell n\left(x+\frac{1}{2}+\sqrt{x^2+x+1}\right)+C \\ & \alpha=\frac{7}{2} \\ & \alpha+2 \beta=16 \end{aligned}$$

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