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JEE Main 2021 · Mathematics · Indefinite and Definite Integrals · medium

The value of ∫0^1 xtan ^-1x dx is

Single correct option

The value of ∫01xtan⁡−1x dx\displaystyle\int_0^1 x\tan^{-1}x\,dx is
  1. A. π4\dfrac{\pi}{4}
  2. B. π−24\dfrac{\pi-2}{4}
  3. C. π−14\dfrac{\pi-1}{4}
  4. D. π8−12\dfrac{\pi}{8}-\dfrac12
Show answer and solution

Answer: (B) π−24\dfrac{\pi-2}{4}

Integrate by parts with u=tan⁡−1xu=\tan^{-1}x and dv=x dxdv=x\,dx: I=x22tan⁡−1x∣01−12∫01x21+x2dx=π8−12∫01(1−11+x2)dx.I=\frac{x^2}{2}\tan^{-1}x\Big|_0^1-\frac12\int_0^1\frac{x^2}{1+x^2}dx=\frac{\pi}{8}-\frac12\int_0^1\left(1-\frac{1}{1+x^2}\right)dx. The remaining integral equals 1−tan⁡−11=1−π41-\tan^{-1}1=1-\dfrac\pi4. So I=π8−12+π8=π4−12=π−24.I=\frac\pi8-\frac12+\frac\pi8=\frac\pi4-\frac12=\frac{\pi-2}{4}. The distractor π8−12\dfrac\pi8-\dfrac12 results from dropping the second π8\dfrac\pi8 when the sign of the subtracted integral is mishandled. Numerically, π−24≈0.2854\dfrac{\pi-2}{4}\approx0.2854.
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