1. **State the problem:** We need to find the limit $$\lim_{x \to +\infty} \frac{\ln(x^2 + 3x + 2) - \ln(x^2 - 4x + 2)}{x^2}.$$\n\n2. **Use logarithm properties:** Recall that $\ln a - \ln b = \ln \frac{a}{b}$. So rewrite the expression as $$\lim_{x \to +\infty} \frac{\ln \left( \frac{x^2 + 3x + 2}{x^2 - 4x + 2} \right)}{x^2}.$$\n\n3. **Analyze the fraction inside the logarithm:** For large $x$, the dominant terms in numerator and denominator are $x^2$. Factor $x^2$ out: $$\frac{x^2 + 3x + 2}{x^2 - 4x + 2} = \frac{x^2(1 + \frac{3}{x} + \frac{2}{x^2})}{x^2(1 - \frac{4}{x} + \frac{2}{x^2})} = \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}}.$$\n\n4. **Simplify the limit inside the logarithm as $x \to +\infty$:** $$\lim_{x \to +\infty} \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}} = \frac{1 + 0 + 0}{1 - 0 + 0} = 1.$$\n\n5. **Use the fact that $\ln(1) = 0$:** So the numerator $\ln \left( \frac{x^2 + 3x + 2}{x^2 - 4x + 2} \right) \to 0$ as $x \to +\infty$.\n\n6. **Apply the limit to the whole expression:** The numerator tends to 0, denominator tends to $+\infty$ (since $x^2 \to +\infty$), so the whole fraction tends to 0.\n\n7. **More rigor with expansion:** Use the expansion for $\ln(1 + u) \approx u$ when $u \to 0$. Define $$u = \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}} - 1 = \frac{(1 + \frac{3}{x} + \frac{2}{x^2}) - (1 - \frac{4}{x} + \frac{2}{x^2})}{1 - \frac{4}{x} + \frac{2}{x^2}} = \frac{\frac{3}{x} + \frac{4}{x}}{1 - \frac{4}{x} + \frac{2}{x^2}} = \frac{\frac{7}{x}}{1 - \frac{4}{x} + \frac{2}{x^2}}.$$\n\nAs $x \to +\infty$, denominator tends to 1, so $$u \approx \frac{7}{x}.$$\n\nTherefore, $$\ln \left( \frac{x^2 + 3x + 2}{x^2 - 4x + 2} \right) \approx \frac{7}{x}.$$\n\n8. **Substitute back into the limit:** $$\lim_{x \to +\infty} \frac{\ln \left( \frac{x^2 + 3x + 2}{x^2 - 4x + 2} \right)}{x^2} \approx \lim_{x \to +\infty} \frac{\frac{7}{x}}{x^2} = \lim_{x \to +\infty} \frac{7}{x^3} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$