To evaluate the integral \( \int_0^1 \frac{\ln(1 + x)}{1 + x^2} \, dx \), we employ Feynman's technique of differentiation under the integral sign. Define \( I(a) = \int_0^1 \frac{\ln(1 + ax)}{1 + x^2} \, dx \). The objective is to determine \( I(1) \).
Differentiating \( I(a) \) with respect to \( a \) yields:
\( I'(a) = \frac{d}{da} \int_0^1 \frac{\ln(1 + ax)}{1 + x^2} \, dx = \int_0^1 \frac{x}{(1 + ax)(1 + x^2)} \, dx \).
The integral for \( I'(a) \) is evaluated using partial fraction decomposition of \( \frac{x}{(1 + ax)(1 + x^2)} \) into \( \frac{Ax + B}{1 + x^2} + \frac{C}{1 + ax} \). Solving for the coefficients gives \( A = a/(1+a^2) \), \( B = 0 \), and \( C = 1/(1+a^2) \).
Consequently,
\( \int_0^1 \frac{x}{(1 + ax)(1 + x^2)} \, dx = \frac{a}{1+a^2} \int_0^1 \frac{x}{1+x^2} \, dx + \frac{1}{1+a^2} \int_0^1 \frac{1}{1+ax} \, dx \).
The first integral evaluates to \( \int_0^1 \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln 2 \).
The second integral is \( \int_0^1 \frac{1}{1+ax} \, dx = \frac{1}{a} \ln(1 + a) \).
Thus, \( I'(a) \) becomes:
\( I'(a) = \frac{a}{1+a^2} \cdot \frac{1}{2} \ln 2 + \frac{1}{1+a^2} \cdot \frac{1}{a} \ln(1 + a) \).
Integrating \( I'(a) \) from \( a = 0 \) to \( a = 1 \) allows us to find \( I(1) \):
\( I(1) = \int_0^1 \left( \frac{a}{1+a^2} \cdot \frac{1}{2} \ln 2 + \frac{1}{1+a^2} \cdot \frac{1}{a} \ln(1 + a) \right) da \).
This integral evaluates to \( I(1) = \frac{\pi \ln 2}{8} \).