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MCQMediumJEE Main 2023 · 29 January, Shift 2Definite Integrals

Mathematics Question from JEE Main 2023 · 29 January, Shift 2

The value of the integral ∫12t4+1t6+1 dt\int_{1}^2 \frac{t^4 + 1}{t^6 + 1} \, dt is:

  • A

    tan⁡−1(12)+13tan⁡−1(8)−π3\tan^{-1}\left(\frac{1}{2}\right) + \frac{1}{3}\tan^{-1}\left(8\right) - \frac{\pi}{3}

  • B

    tan⁡−1(2)−13tan⁡−1(8)+π3\tan^{-1}\left({2}\right) - \frac{1}{3}\tan^{-1}\left({8}\right) + \frac{\pi}{3}

  • C

    tan⁡−1(2)+13tan⁡−1(8)−π3\tan^{-1}\left({2}\right) + \frac{1}{3}\tan^{-1}\left({8}\right) - \frac{\pi}{3}

  • D

    tan⁡−1(12)−13tan⁡−1(8)+π3\tan^{-1}\left(\frac{1}{2}\right) - \frac{1}{3}\tan^{-1}\left({8}\right) + \frac{\pi}{3}

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