Question

$\underset{0}{\overset{1}{\int }}\frac{\log (1+x)}{1+x^2}dx$

• A. $\frac{1}{2}$
• B. $\frac{\pi }{8}\log 2$
• C. $\frac{\pi }{2}\log 2$
• D. $\frac{\pi }{4}\log 2$

Answer 1

Answer: (B)

By substituting $x=\tan \theta$
$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan \theta )d\theta$
$=\frac{\pi }{8}\log 2$