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MCQMediumJEE Main 2024 · 27 January, Shift 2Indefinite Integrals

Mathematics Question from JEE Main 2024 · 27 January, Shift 2

The integral ∫x8−x2(x12+3x6+1)arctan⁡(x3+1x3) dx\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \arctan\left(x^3 + \frac{1}{x^3}\right)} \, dx is equal to:

  • A

    log⁡(arctan⁡(x3+1x3)1/3)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^{1/3}\right) + C

  • B

    log⁡(arctan⁡(x3+1x3)1/2)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^{1/2}\right) + C

  • C

    log⁡(arctan⁡(x3+1x3))+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)\right) + C

  • D

    log⁡(arctan⁡(x3+1x3)3)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^3\right) + C

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