Find the value of \(\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x\) is

Let \(I=\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x\)
Put \(x=\tan \theta\)
\(\Rightarrow \quad d x=\sec ^{2} \theta d \theta\)
When \(x=0, \theta=0\)
and when \(x=1, \theta=\pi / 4\)
\(\begin{aligned} \therefore \quad \int_{0}^{\pi / 4} & \frac{\log (1+\tan \theta)}{1+\tan ^{2} \theta}\left(\sec ^{2} \theta\right) d \theta \\ I &=\int_{0}^{\pi / 4} \log (1+\tan \theta) d \theta ...(i) \\=& \int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta \\=& \int_{0}^{\pi / 4} \log \left[1+\frac{\tan \pi / 4-\tan \theta}{1+\tan \pi / 4 \tan \theta}\right] d \theta \\ &=\int_{0}^{\pi / 4} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta \\=& \int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan \theta}\right) d \theta \end{aligned}\)
\(\int_{0}^{\pi / 4}[\log 2-\log (1+\tan \theta)] d \theta...(ii)\)
On adding Eqs. (i) and (ii), we get
\(\begin{aligned} 2 I &=\int_{0}^{\pi / 4} \log 2 d \theta \\ &=\log 2 \int_{0}^{\pi / 4} \operatorname{ld\theta }=\log (\theta)_{0}^{\pi / 4} \\ &=\log (\pi / 4-0)=\frac{\pi}{4} \log 2 \\ I &=\frac{\pi}{8} \log 2 \end{aligned}\)