The value of integral $\int\limits_0^1 \frac{\log (1+x)}{1+x^2} d x$, is
Answer & explanation
Correct answer: option 1
Let $I=\int\limits_0^1 \frac{\log (1+x)}{1+x^2} d x$
Putting $x=\tan \theta$, we get
$I=\int\limits_0^{\pi / 4} \log (1+\tan \theta) d \theta=\frac{\pi}{8} \log _e 2$