Question

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

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

Answer 1

Answer: (D)

$I=\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$
Let $x=\tan \theta$
$\Rightarrow d x=\sec ^{2} \theta d \theta$
$I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta$
$=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan \theta}\right) d \theta$
$=\int_{0}^{\frac{\pi}{4}} \log 2 d \theta-\int_{0}^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\Rightarrow 2 I=\frac{\pi}{4} \log 2$
$\Rightarrow I=\frac{\pi}{8} \log 2$