${\int }_0^1\frac{\ln (1+x)}{1+x^2}dx$

Solution

1. Let $x=\tan \theta$, so $dx={\sec }^2\theta d\theta$. The integral becomes $I={\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta$.
2. Apply the property ${\int }_0^{a}f(\theta )d\theta ={\int }_0^{a}f(a-\theta )d\theta$ to get $I={\int }_0^{\pi /4}\ln (1+\tan (\frac{\pi }{4}-\theta ))d\theta$.
3. Simplify the integrand using the identity $\tan (\frac{\pi }{4}-\theta )=\frac{1-\tan \theta }{1+\tan \theta }$, resulting in $I={\int }_0^{\pi /4}\ln (\frac{2}{1+\tan \theta })d\theta$.
4. Expand the logarithm to obtain $I={\int }_0^{\pi /4}\ln 2d\theta -{\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta$, which simplifies to $I=\frac{\pi }{4}\ln 2-I$.
5. Solve for $I$ to find $2I=\frac{\pi }{4}\ln 2$, which yields $I=\frac{\pi }{8}\ln 2$.
6. Hence the answer is (A).