1. **State the problem:** We need to evaluate the definite integral $$\int_1^4 x^2 \ln x \, dx$$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = \ln x \quad \Rightarrow \quad du = \frac{1}{x} dx$$ $$dv = x^2 dx \quad \Rightarrow \quad v = \frac{x^3}{3}$$ 3. **Apply integration by parts:** $$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx$$ 4. **Integrate remaining integral:** $$\int x^2 dx = \frac{x^3}{3}$$ 5. **Substitute back:** $$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$$ 6. **Evaluate definite integral from 1 to 4:** $$\left[ \frac{x^3}{3} \ln x - \frac{x^3}{9} \right]_1^4 = \left( \frac{4^3}{3} \ln 4 - \frac{4^3}{9} \right) - \left( \frac{1^3}{3} \ln 1 - \frac{1^3}{9} \right)$$ 7. **Calculate values:** $$\ln 1 = 0$$ $$\frac{4^3}{3} = \frac{64}{3}$$ $$\frac{4^3}{9} = \frac{64}{9}$$ So, $$= \frac{64}{3} \ln 4 - \frac{64}{9} - \left(0 - \frac{1}{9} \right) = \frac{64}{3} \ln 4 - \frac{64}{9} + \frac{1}{9} = \frac{64}{3} \ln 4 - \frac{63}{9} = \frac{64}{3} \ln 4 - 7$$ **Final answer:** $$\boxed{\frac{64}{3} \ln 4 - 7}$$