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∫ e^2x - 1 e^2x + 1 dx =

CUET UG 2025 — Mathematics Calculus

2025mcqmedium

∫e2x−1e2x+1dx=\int \frac{e^{2x} - 1}{e^{2x} + 1} dx =∫e2x+1e2x−1​dx=

Official previous-year question

Held on 27 May 2025 · Verified 13 Jul 2026.

Options

  1. A

    log⁡∣ex+e−x∣+C\log|e^x + e^{-x}| + Clog∣ex+e−x∣+C : CCC is an arbitrary constant

  2. B

    log⁡∣e2x+1∣+C\log|e^{2x} + 1| + Clog∣e2x+1∣+C : CCC is an arbitrary constant

  3. C

    log⁡∣e2x−e−x∣+C\log|e^{2x} - e^{-x}| + Clog∣e2x−e−x∣+C : CCC is an arbitrary constant

  4. D

    log⁡∣e2x−1∣+C\log|e^{2x} - 1| + Clog∣e2x−1∣+C : CCC is an arbitrary constant

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