Question

${\int }_0^1\frac{\log (1+x)}{1+x^2}d{x}_{=}$

• A. $\pi$ log2
• B. $\frac{\pi }{8}$ log2
• C. $\frac{\pi }{4}$ log2
• D. $-\pi$ log2

Answer 1

The correct option is B $\frac{\pi }{8}$ log2

${\int }_0^1\frac{log(1+x)}{1+x^2}dx$
$x=tan\theta$
$I={\int }_0^{\frac{\pi }{4}}log(1+tan\theta )d\theta$
$I={\int }_0^{\frac{\pi }{4}}log(1+\frac{1-tan\theta }{1+tan\theta })d\theta$
$I={\int }_0^{\frac{\pi }{4}}log\left(\frac{2}{1+tan\theta }\right)d\theta$
$I={\int }_0^{\frac{\pi }{4}}log2-1$
$2I=\frac{\pi }{4}log2$
$I=\frac{\pi }{8}log2$