Question

$\int\limits _{0}^{1} \frac {8 \log (1+x)}{1+x^2}dx$ is equal to

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

Answer 1

Answer: (A)

$\pi log 2$

Answer 2

Let $I=\int\limits_{0}^{1} \frac{8}{1+x^{2}} log \left(1 + x \right)dx$. Put $x = tan \, \theta$ $\therefore I=8 \int\limits_{0}^{\pi /4} \frac{log\left(1+tan\,\theta\right)sec^{2}\,\theta}{1+tan^{2}\,\theta}$ $=8 \int\limits_{0}^{\pi /4}log\left(1+tan\,\theta\right)d\theta$ Also , $I=8 \int\limits_{\pi /4}^{\pi /4} log \left[1+tan\left(\frac{\pi}{4}-\theta\right)\right]d \theta$ $=8 \int\limits_{0}^{\pi /4} log\left(1+\frac{1-tan\,\theta}{1+tan\,\theta}\right)d \theta$ $=8 \int\limits_{0}^{\pi/ 4} log\left(\frac{2}{1+tan\,\theta}\right)dx$ $=8\left[log\,2 \int\limits_{0}^{\pi /4} d\theta\right] \Rightarrow2I$ $=log\,2\cdot\left(\frac{\pi}{4}\right)8$ $\Rightarrow I=\pi \, log\,2=2\pi\, log\,2$