PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 2 · Mathematics · Indefinite and Definite Integrals · medium

The value of ∫0^1 (ln (1+x))/(1+x^2 dx is

Single correct option

The value of ∫01ln⁡(1+x)1+x2 dx\displaystyle\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx is
  1. A. π4ln⁡2\dfrac{\pi}{4}\ln 2
  2. B. π2ln⁡2\dfrac{\pi}{2}\ln 2
  3. C. π8ln⁡2\dfrac{\pi}{8}\ln 2
  4. D. π16ln⁡2\dfrac{\pi}{16}\ln 2
Show answer and solution

Answer: (C) π8ln⁡2\dfrac{\pi}{8}\ln 2

Put x=tan⁡θx=\tan\theta, so I=∫0π/4ln⁡(1+tan⁡θ) dθI=\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta. Apply the property ∫0af(θ) dθ=∫0af(a−θ) dθ\int_0^a f(\theta)\,d\theta=\int_0^a f(a-\theta)\,d\theta with a=π/4a=\pi/4. Since 1+tan⁡(π/4−θ)=21+tan⁡θ1+\tan(\pi/4-\theta)=\dfrac{2}{1+\tan\theta}, we get I=∫0π/4ln⁡21+tan⁡θ dθ=π4ln⁡2−II=\int_0^{\pi/4}\ln\dfrac{2}{1+\tan\theta}\,d\theta=\dfrac{\pi}{4}\ln2-I. Hence 2I=π4ln⁡22I=\dfrac{\pi}{4}\ln 2 and I=π8ln⁡2≈0.2722I=\dfrac{\pi}{8}\ln 2\approx0.2722, which agrees with numerical quadrature. The option π4ln⁡2\frac{\pi}{4}\ln2 is the value of 2I2I and comes from forgetting to halve; the others are off by further factors of 2.
Report a mistake

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake