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NDA 2020 Maths · Mathematics · Integrals · medium

The value of the definite integral ∫0^1tan ^-1x dx, found by parts, is

Single correct option

The value of the definite integral ∫01tan⁡−1x dx\displaystyle\int_0^1\tan^{-1}x\,dx, found by parts, is
  1. A. π4+12ln⁡2\dfrac\pi4+\dfrac12\ln2
  2. B. π4\dfrac\pi4
  3. C. π4−12ln⁡2\dfrac\pi4-\dfrac12\ln2
  4. D. π4−ln⁡2\dfrac\pi4-\ln2
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Answer: (C) π4−12ln⁡2\dfrac\pi4-\dfrac12\ln2

Integrate by parts with u=tan⁡−1xu=\tan^{-1}x, dv=dxdv=dx: ∫tan⁡−1x dx=xtan⁡−1x−∫x1+x2dx=xtan⁡−1x−12ln⁡(1+x2)\displaystyle\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\dfrac{x}{1+x^2}dx=x\tan^{-1}x-\dfrac12\ln(1+x^2). Evaluating between 00 and 11 gives π4−12ln⁡2\dfrac\pi4-\dfrac12\ln2. The option π4−ln⁡2\dfrac\pi4-\ln2 forgets the factor 12\dfrac12 from ∫2x1+x2\int\dfrac{2x}{1+x^2}, and the option with ++ misses the minus sign of the by-parts formula.
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Original question written in the NDA 2020 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