QJEEM25Ma624e: JEE Main 2025 Mathematics - Class 12 JEEM25Ma624e Solution: If , where is the constant of integration, then is equal to…

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

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