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NDA 2025 Maths · Mathematics · Integrals · hard

Find the indefinite integral: ∫(dx)/(e^x+e^-x)

Single correct option

Find the indefinite integral: ∫dxex+e−x\displaystyle\int\dfrac{dx}{e^x+e^{-x}}
  1. A. ln⁡(ex+e−x)+C\ln(e^x+e^{-x})+C
  2. B. tan⁡−1(ex)+C\tan^{-1}(e^x)+C
  3. C. tan⁡−1(e−x)+C\tan^{-1}(e^{-x})+C
  4. D. 12ln⁡(e2x+1)+C\dfrac12\ln(e^{2x}+1)+C
Show answer and solution

Answer: (B) tan⁡−1(ex)+C\tan^{-1}(e^x)+C

Multiply numerator and denominator by exe^x: ∫exe2x+1dx\int\frac{e^x}{e^{2x}+1}dx. Putting u=exu=e^x gives ∫du1+u2=tan⁡−1(ex)+C\int\frac{du}{1+u^2}=\tan^{-1}(e^x)+C. The logarithm option would be the integral of ex−e−xex+e−x\frac{e^x-e^{-x}}{e^x+e^{-x}}, and tan⁡−1(e−x)\tan^{-1}(e^{-x}) has the wrong sign of the exponent.
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Original question written in the NDA 2025 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake