Question

The value of ${\int }_0^1\frac{\log (1+x)}{1+x^2}dx$ is?

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

Answer 1

The correct option is A $\frac{\pi }{8}{\log }_{e}2$

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

Putting, $x=tan\theta$

$dx=se{c}^2\theta d\theta =(1+x^2)d\theta$

Now we have, $I={\int }_0^{\pi /4}log(1+tan\theta )d\theta$

$I={\int }_0^{\pi /4}log(1+tan(\frac{\pi }{4}-\theta \left)\right)d\theta$ (Using the property of integration)

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

$I=\frac{\pi }{8}{\log }_{e}2$