JEE Main 2025 Apr 7 Shift 2, Mathematics Q22: Methods of Integration

JEE Main2025Apr 7, Shift 2Mathematics

If $\int (\frac{1}{x ^{2}} + \frac{1}{x ^{4}}) (23 3 x^{- 1} + x^{- 3}) d x = - \frac{α}{3 ( α + 1 )} (3 x^{β} + x^{γ})^{\frac{α + 1}{α}} + C, x > 0,$

$(α, β, γ \in Z)$ , where C is the constant of integration, then $α + β + γ$ is equal to _____.

Your answer

Solution

Let $t = 3 x^{- 1} + x^{- 3}$ , so $\frac{d t}{d x} = - 3 x^{- 2} - 3 x^{- 4} = - 3 (\frac{1}{x ^{2}} + \frac{1}{x ^{4}})$ , i.e. $(\frac{1}{x ^{2}} + \frac{1}{x ^{4}}) d x = - \frac{d t}{3}$ .

The integral becomes $\int t^{1/23} (- \frac{d t}{3}) = - \frac{1}{3} \cdot \frac{t ^{24/23}}{24/23} + C = - \frac{23}{72} t^{24/23} + C$ .

Rewriting the coefficient: $- \frac{23}{72} = - \frac{23}{3 ( 24 )} = - \frac{α}{3 ( α + 1 )}$ with $α = 23$ (so that $α + 1 = 24$ ), and the exponent $\frac{24}{23} = \frac{α + 1}{α}$ matches consistently.

Comparing $t^{24/23} = (3 x^{- 1} + x^{- 3})^{24/23}$ with $(3 x^{β} + x^{γ})^{(α + 1) / α}$ gives $β = - 1, γ = - 3$ .

$∴ α + β + γ = 23 - 1 - 3 = 19$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q22: Methods of Integration

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