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If ∫ (e^2x+2e^x-e^-x-1)e^(e^x+e^-x)dx=g(x)e^(e^x+e^-x)+c, where c is a constant of integration, then g(0) is

JEE Main 2020 — Mathematics Calculus

2020mcqhard

If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c,\displaystyle \int ({e}^{2x}+2{e}^{x}-{e}^{-x}-1){e}^{({e}^{x}+{e}^{-x})}dx=g(x){e}^{({e}^{x}+{e}^{-x})}+c,∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c, where ccc is a constant of integration, then g(0)g(0)g(0) is

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

  1. A

    eee

  2. B

    e2{e}^{2}e2

  3. C

    111

  4. D

    222

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