Given integral
\(I=\int_0^1 \frac{\log _e(1+x)}{1+x^2} d x\)
put \(x=\tan \theta\), so at \(x=0, \theta=0\) and
at \(x=1, \theta=\frac{\pi}{4}\)
and \(d x=\sec ^2 \theta d \theta\)
Now, \(I=\int_0^{\frac{\pi}{4}} \frac{\log _e(1+\tan \theta)}{\sec ^2 \theta} \sec ^2 \theta d \theta\)
\(=\int_0^{\frac{\pi}{4}} \log _e(1+\tan \theta) d \theta\) ...(i)
On, applying property, \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\), we get
\(\begin{aligned}
& I=\int_0^{\frac{\pi}{4}} \log _e\left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta \\
& =\int_0^{\frac{\pi}{4}} \log _e\left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta \\
& =\int_0^{\frac{\pi}{4}} \log _e\left(\frac{2}{1+\tan \theta}\right) d \theta \\
& =\int_0^{\frac{\pi}{4}} \log _e(2) d x-\int_0^{\frac{\pi}{4}} \log _e(1+\tan \theta) d \theta \\
& \Rightarrow \quad I=\frac{\pi}{4} \log _e(2)-I \quad \text{[from Eq. (i)]} \\
& \Rightarrow \quad 2 I=\frac{\pi}{4} \log _e(2) \Rightarrow I=\frac{\pi}{8} \log _e(2) \\
\end{aligned}\)
Hence, option (4) is correct.