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Friday, 9 October

JEE Main 2025 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 7 April 2025, Shift 2, Maths Q24

Question 24 of 75 in this shift, Maths question 24 of 25, Section B.

If ∫(1x+1x3)3x−24+x−2623 dx=−α3(α+1)(3xβ+xγ)α+1α+C\int\left(\frac{1}{x}+\frac{1}{x^3}\right)\sqrt[23]{3x^{-24}+x^{-26}}\,dx=-\frac{\alpha}{3(\alpha+1)}\left(3x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C, x>0x>0, (α,β,γ∈Z)(\alpha,\beta,\gamma\in\mathbf{Z}), where CC is the constant of integration, then α+β+γ\alpha+\beta+\gamma is equal to ________.

Official answer

19

NTA final key.

Chapter
Integrals
Topic
Integration by Substitution
Idea tested
Factoring Highest Power for Substitution

Same idea in other shifts

Asked 2× in all
  1. 8 Jan 2020, Shift 1 · Q60If ∫cos⁡x dxsin⁡3x (1+sin⁡6x)2/3=f(x) (1+sin⁡6x)1/λ+c\int\dfrac{\cos x\,dx}{\sin^3 x\,(1+\sin^6 x)^{2/3}}=f(x)\,(1+\sin^6 x)^{1/\lambda}+c where c is a constant of integration, then…HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.