Question

The value of $\underset{0}{\overset{1}{\int }}\frac{8\log (1+x)}{1+x^2}dx$ 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,
$I=\underset{0}{\overset{1}{\int }}\frac{8\log (1+x)}{1+x^2}dx$
Put $x=\tan \theta$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}8\cdot \frac{\log (1+\tan \theta )}{{\sec }^2\theta }{\sec }^2\theta d\theta$
$=8\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan \theta )d\theta$
$=8\underset{0}{\overset{\pi /4}{\int }}\log \left(1+\frac{1-\tan \theta }{1+\tan \theta }\right)d\theta$
$=8\underset{0}{\overset{\pi /4}{\int }}\log \left(\frac{2}{1+\tan \theta }\right)d\theta$
$=8(\log 2)\cdot \frac{\pi }{4}-I$
$\Rightarrow 2I=2\pi \log 2$
$\Rightarrow I=\pi \log 2$