Question

The value of ${\int }_0^1\frac{8\log (1+x)}{1+x^2}$ dx is

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

Answer 1

The correct option is A $\pi \log 2$

$I={\int }_0^1\frac{8\log (1+x)}{1+x^2}dx$
Let $x=\tan \theta =$ dx $={\sec }^2\theta d\theta$
$I={\int }_0^{\pi /4}8\log (1+\tan \theta )d\theta$
$I=8{\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-\theta \left)\right)d\theta$
$=8{\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan \theta }\right)d\theta$
$=8{\int }_0^{\pi /4}(\log 2-\log (1+\tan \theta \left)\right)d\theta$
$I=4{\int }_0^{\pi /4}\log 2d\theta =\pi log2$
Hence, option 'A' is correct.