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Friday, 9 October

JEE Main 2025 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 4 April 2025, Shift 2, Maths Q25

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

If ∫(1+x2+x)10(1+x2−x)9dx=1m((1+x2+x)n(n1+x2−x))+C\int\frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^{9}}dx=\frac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^{n}\left(n\sqrt{1+x^2}-x\right)\right)+C where C is the constant of integration and m,n∈Nm,n\in\mathbf{N}, then m+nm+n is equal to __________.

Official answer

379

NTA final key.

Chapter
Integrals
  1. 5 Apr 2026, Shift 1 · Q20The value of the integral ∫π/6π/3(4−cosec⁡2xcos⁡4x)dx\int_{\pi/6}^{\pi/3}\left(\frac{4 - \operatorname{cosec}^2 x}{\cos^4 x}\right)dx is:MediumSingle correct
  2. 5 Apr 2026, Shift 2 · Q18Let (21−a+21+a)(2^{1-a}+2^{1+a}), f(a)f(a), (3a+3−a)(3^a+3^{-a}) be in A.P. and α\alpha be the minimum value of f(a)f(a). Then the value of the integral…HardSingle correct
  3. 3 Apr 2025, Shift 1 · Q18Let f(x)=∫x33−x2 dxf(x) = \int x^3\sqrt{3 - x^2}\,dx. If 5f(2)=−45f(\sqrt{2}) = -4, then f(1)f(1) is equal toMediumSingle correct
  4. 7 Apr 2025, Shift 2 · Q24If…HardNumerical value
  5. 6 Apr 2024, Shift 2 · Q12If ∫1a2sin⁡2x+b2cos⁡2xdx=112tan⁡−1(3tan⁡x)+constant\int\frac{1}{a^2\sin^2x+b^2\cos^2x}dx=\frac{1}{12}\tan^{-1}(3\tan x)+\text{constant}, then the maximum value of asin⁡x+bcos⁡xa\sin x+b\cos x, is :MediumSingle correct
  6. 8 Apr 2024, Shift 1 · Q10Let I(x)=∫6sin⁡2x(1−cot⁡x)2dxI(x)=\int\frac{6}{\sin^2x(1-\cot x)^2}dx. If I(0)=3I(0)=3, then I(π12)I\left(\frac{\pi}{12}\right) is equal toMediumSingle 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.