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

Indefinite Integrals — JEE Main · 2025 · 24 Jan 2025 · Shift 2 Question with Solution

\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} x + \frac{1}{2} + x^{2} + x + 1 + C

, where C is the constant of integration, then

α + 2 β

is equal to ____

Answer: 16

Write the numerator in terms of

x^{2} + x + 1

and its derivative

2 x + 1

2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B (2 x + 1) + C .

Expanding and comparing coefficients:

x^{2} : A = 2, x : A + 2 B = 5 \Rightarrow B = \frac{3}{2}, const : A + B + C = 9 \Rightarrow C = \frac{11}{2} .

So the integral splits as

2 \int x^{2} + x + 1 d x + \frac{3}{2} \int \frac{2 x + 1}{x ^{2} + x + 1} d x + \frac{11}{2} \int \frac{d x}{x ^{2} + x + 1} .

The middle integral is direct:

\frac{3}{2} \int \frac{2 x + 1}{x ^{2} + x + 1} d x = 3 x^{2} + x + 1

.

Completing the square,

x^{2} + x + 1 = (x + \frac{1}{2})^{2} + \frac{3}{4}

. Using the standard forms

\int u^{2} + k^{2} d u = \frac{u}{2} u^{2} + k^{2} + \frac{k ^{2}}{2} ln u + u^{2} + k^{2}, \int \frac{d u}{u ^{2} + k ^{2}} = ln u + u^{2} + k^{2},

with

u = x + \frac{1}{2}

k^{2} = \frac{3}{4}

2 \int x^{2} + x + 1 d x = (x + \frac{1}{2}) x^{2} + x + 1 + \frac{3}{4} ln x + \frac{1}{2} + x^{2} + x + 1,

\frac{11}{2} \int \frac{d x}{x ^{2} + x + 1} = \frac{11}{2} ln x + \frac{1}{2} + x^{2} + x + 1 .

Combining everything:

x x^{2} + x + 1 + \frac{7}{2} x^{2} + x + 1 + \frac{25}{4} ln x + \frac{1}{2} + x^{2} + x + 1 + C .

Matching with the given form,

α = \frac{7}{2}, β = \frac{25}{4}

, so

α + 2 β = \frac{7}{2} + \frac{25}{2} = 16.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer