Question

The value of $\int\limits_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$ is

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

Answer 1

Answer: (B)

We have,
$I=\int\limits_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$
Put $x=\tan \theta$
$\Rightarrow I=\int\limits_{0}^{\frac{\pi}{4}} 8 \cdot \frac{\log (1+\tan \theta)}{\sec ^{2} \theta} \sec ^{2} \theta d \theta$
$=8 \int\limits_{0}^{\pi / 4} \log (1+\tan \theta) d \theta$
$=8 \int\limits_{0}^{\pi / 4} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta$
$=8 \int\limits_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan \theta}\right) d \theta$
$=8(\log 2) \cdot \frac{\pi}{4}-I$
$\Rightarrow 2 I=2 \pi \log 2$
$\Rightarrow I=\pi \log 2$