1. **State the problem:** Find the exact value of the integral $$\int_1^2 \ln(3x) \, dx$$ in the form $a + \ln b$, where $a$ and $b$ are integers. 2. **Recall the integration formula:** For integration by parts, $$\int u \, dv = uv - \int v \, du$$. 3. **Set up integration by parts:** Let $$u = \ln(3x)$$ and $$dv = dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} \, dx$$ (using chain rule) - $$v = x$$ 5. **Apply integration by parts:** $$\int \ln(3x) \, dx = x \ln(3x) - \int x \cdot \frac{1}{x} \, dx = x \ln(3x) - \int 1 \, dx = x \ln(3x) - x + C$$ 6. **Evaluate definite integral from 1 to 2:** $$\int_1^2 \ln(3x) \, dx = \left[ x \ln(3x) - x \right]_1^2 = (2 \ln(6) - 2) - (1 \ln(3) - 1)$$ 7. **Simplify the expression:** $$= 2 \ln(6) - 2 - \ln(3) + 1 = (2 \ln(6) - \ln(3)) - 1$$ 8. **Use logarithm properties:** $$2 \ln(6) - \ln(3) = \ln(6^2) - \ln(3) = \ln(36) - \ln(3) = \ln\left(\frac{36}{3}\right) = \ln(12)$$ 9. **Final answer:** $$\int_1^2 \ln(3x) \, dx = \ln(12) - 1$$ This matches the form $a + \ln b$ with $a = -1$ and $b = 12$.