JEE Main 2025 Jan 24 Shift 2, Mathematics Q25: Methods of Integration

Express the numerator as a combination of (i) the denominator's argument, (ii) its derivative, and (iii) a constant:

$2x^2+5x+9=A(x^2+x+1)+B(2x+1)+C_0$

Compare coefficients: $A=2$; $A+2B=5\Rightarrow B=3/2$; $A+B+C_0=9\Rightarrow C_0=9-2-3/2=11/2$.

So the integral splits as:

$I=2\int \sqrt{x^2+x+1}dx+\frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{11}{2}\int \frac{dx}{\sqrt{x^2+x+1}}$

Complete the square: $x^2+x+1={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}$. Let $u=x+\frac{1}{2}$, $k^2=3/4$.

Standard integrals:

$\int \sqrt{u^2+k^2}du=\frac{u}{2}\sqrt{u^2+k^2}+\frac{k^2}{2}\ln \left|u+\sqrt{u^2+k^2}\right|+C$

$\int \frac{du}{\sqrt{u^2+k^2}}=\ln \left|u+\sqrt{u^2+k^2}\right|+C$

$\int \frac{2x+1}{\sqrt{x^2+x+1}}dx=2\sqrt{x^2+x+1}+C$ (the numerator is the derivative of the radicand).

Assembling:

$I=2\left[\frac{u}{2}\sqrt{x^2+x+1}+\frac{3}{8}\ln \left|u+\sqrt{x^2+x+1}\right|\right]+3\sqrt{x^2+x+1}+\frac{11}{2}\ln \left|u+\sqrt{x^2+x+1}\right|$

$=\left(x+\frac{1}{2}\right)\sqrt{x^2+x+1}+3\sqrt{x^2+x+1}+\left(\frac{3}{4}+\frac{11}{2}\right)\ln \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|$

$=x\sqrt{x^2+x+1}+\frac{7}{2}\sqrt{x^2+x+1}+\frac{25}{4}\ln \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C$

Matching: $\alpha =\frac{7}{2}$, $\beta =\frac{25}{4}$.

$\alpha +2\beta =\frac{7}{2}+\frac{25}{2}=\frac{32}{2}=16$.

Answer: 16