Question

The value of ${\int }_0^1\frac{8\log (1+x)}{1+x^2}dx$ is

• A. $\left(\frac{\mathrm{\pi }}{8}\right)\log 2$
• B. $\left(\frac{\mathrm{\pi }}{2}\right)\log 2$
• C. $\log 2$
• D. $\pi \log 2$

Answer 1

The correct option is D

$\pi \log 2$

The explanation for the correct option.

Evaluate the integral.

$$\begin{gathered} {\int }_0^1\frac{8\log (1+x)}{1+x^2}dx \\ x=\tan \theta \\ se{c}^2\theta d\theta =dx \\ \theta =0,x=0,\theta =\frac{\mathrm{\pi }}{4} \\ I=8{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\log (1+\tan \theta )}{1+{\tan }^2\theta }se{c}^2\theta d\theta \\ =8{\int }_0^{\frac{\mathrm{\pi }}{4}}\log (1+\tan \theta )d\theta \\ =8{\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(\frac{2}{1+\tan \theta }\right)d\theta \\ =8{\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(2\right)-\log (1+\tan \theta )d\theta \\ =8\log 2\times \frac{\mathrm{\pi }}{4} \\ =\mathrm{\pi }\log 2 \end{gathered}$$

Hence, option D is correct .