The value of integral $\int\limits_0^1 \frac{\log (1+x)}{1+x^2} d x$, is

$\frac{\pi}{8} \log _e 2$ $\frac{\pi}{4} \log _e 2$ $-\frac{\pi}{8} \log _e 2$ $-\frac{\pi}{4} \log _e 2$

$\frac{\pi}{8} \log _e 2$

Let $I=\int\limits_0^1 \frac{\log (1+x)}{1+x^2} d x$ Putting $x=\tan \theta$, we get $I=\int\limits_0^{\pi / 4} \log (1+\tan \theta) d \theta=\frac{\pi}{8} \log _e 2$