1. **State the problem:** Evaluate the definite integral $$\int_1^4 x^2 \ln(x) \, dx$$. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = \ln(x) \implies du = \frac{1}{x} dx$$ $$dv = x^2 dx \implies v = \frac{x^3}{3}$$ 4. **Apply integration by parts:** $$\int_1^4 x^2 \ln(x) \, dx = \left. \frac{x^3}{3} \ln(x) \right|_1^4 - \int_1^4 \frac{x^3}{3} \cdot \frac{1}{x} dx$$ 5. **Simplify the integral:** $$= \left. \frac{x^3}{3} \ln(x) \right|_1^4 - \frac{1}{3} \int_1^4 x^2 dx$$ 6. **Evaluate the remaining integral:** $$\int_1^4 x^2 dx = \left. \frac{x^3}{3} \right|_1^4 = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21$$ 7. **Substitute back:** $$= \left( \frac{4^3}{3} \ln(4) - \frac{1^3}{3} \ln(1) \right) - \frac{1}{3} \times 21$$ 8. **Simplify terms:** Since $$\ln(1) = 0$$, $$= \frac{64}{3} \ln(4) - 0 - 7 = \frac{64}{3} \ln(4) - 7$$ **Final answer:** $$\int_1^4 x^2 \ln(x) \, dx = \frac{64}{3} \ln(4) - 7$$