1. **Problem statement:** We want to evaluate the integral $$I = \int_0^{\infty} \frac{\ln(1 + x^2)}{1 + x^2} \, dx.$$\n\n2. **Key idea:** This integral involves the logarithm and a rational function. A useful substitution is to use the symmetry of the integrand or relate it to a known integral involving the arctangent function.\n\n3. **Step 1: Use substitution** \(x = \tan \theta\), so that \(dx = \sec^2 \theta \, d\theta\) and \(1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta\). The limits change from \(x=0\) to \(\theta=0\), and as \(x \to \infty\), \(\theta \to \frac{\pi}{2}\).\n\n4. **Rewrite the integral:**\n$$I = \int_0^{\pi/2} \frac{\ln(1 + \tan^2 \theta)}{1 + \tan^2 \theta} \cdot \sec^2 \theta \, d\theta = \int_0^{\pi/2} \frac{\ln(\sec^2 \theta)}{\sec^2 \theta} \cdot \sec^2 \theta \, d\theta = \int_0^{\pi/2} \ln(\sec^2 \theta) \, d\theta.$$\n\n5. **Simplify the logarithm:**\n$$\ln(\sec^2 \theta) = 2 \ln(\sec \theta) = 2 \ln \left(\frac{1}{\cos \theta}\right) = -2 \ln(\cos \theta).$$\n\n6. **Integral becomes:**\n$$I = \int_0^{\pi/2} -2 \ln(\cos \theta) \, d\theta = -2 \int_0^{\pi/2} \ln(\cos \theta) \, d\theta.$$\n\n7. **Known integral:** It is a standard result that\n$$\int_0^{\pi/2} \ln(\cos \theta) \, d\theta = -\frac{\pi}{2} \ln 2.$$\n\n8. **Substitute this result:**\n$$I = -2 \left(-\frac{\pi}{2} \ln 2 \right) = \pi \ln 2.$$\n\n**Final answer:**\n$$\boxed{I = \pi \ln 2}.$$