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

Correct answer: 0 aligned & 2 x^2+5 x+9=A(x^2+x+1)+B(2 x+1)+C & A=2 ~B=(3)/(2) C=(11)/(2) & 2 ∫ x^2+x+1 dx+(3)/(2) ∫ 2 x+1 x^2+x+1 dx+(11)/(2) ∫ dx x^2+x+1 & 2 ∫ (x+(1)/(2))^2+( 3 2 )^2 dx+3 x^2+x+1 +(11)/(2) ∫ dx (x+(1)/(2))^2+( 3 2 )^2 & 2( x+(1)/(2) 2 x^2+x+1 +(3)/(8) (x+(1)/(2)+ x^2+x+1 ))+3 x^2+x+1 & +(11)/(2) n(x+(1)/(2)+ x^2+x+1 )+C & α=(7)/(2) β=(25)/(4) \& α+2 β=16 & aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Indefinite Integration

If ∫ 2 x^2+5 x+9 x^2+x+1 ~d x=x x^2+x+1 +α x^2+x+1 +β _e|x+(1)/(2)+...

\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 _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

\qquad

_______.

Correct answer

Step-by-step solution

\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}

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