Question

The value of The value of $\underset{0}{\overset{{}^1}{\int }}\frac{log\left(1+x\right)}{1+x^2}dx$ is

• A. $\frac{\pi }{2}log2$
• B. $\frac{\pi }{4}log2$
• C. $\frac{1}{2}$
• D. $\frac{\pi }{8}log2$

Answer 1

Answer: (D)

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