Q39 / 90 · JEE Main 2011 · Mathematics

The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is

  • A
    $\frac{\pi}{8} \log 2$
  • B
    $\frac{\pi}{2} \log 2$
  • C
    $\log 2$
  • Correct answer
    D
    $\pi \log 2$

Solution

$\mathrm{I}=8 \int_0^1 \frac{\log (1+\mathrm{x})}{1+\mathrm{x}^2} \mathrm{dx}$ $=8 \int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \sec ^2 \theta d \theta(\operatorname{let} x=\tan \theta)$ $=8 \int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log 2 d \theta-8 \int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$ $=8 \log 2 \frac{\pi}{4}-I$ $2 \mathrm{I}=2 \pi \log 2$ $\mathrm{I}=\pi \log 2$

Hence, the correct answer is option: 'D'.

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