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MCQMediumJEE Main 2025 · 23 January, Shift 1Definite Integrals

Mathematics Question from JEE Main 2025 · 23 January, Shift 1

The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{\left( (\log_e x)^2 +1 \right)^{-1}}}{e^{\left( (\log_e x)^2 +1 \right)^{-1}} + e^{\left( (6-\log_e x)^2 +1 \right)^{-1}}} \right) dx is:

  • A

    log⁡2\log 2

  • B

    22

  • C

    11

  • D

    e2e^2

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