1. **State the problem:** We need to solve the integral $$\int x^2 \ln x \, dx$$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ 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:** $$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx$$ 5. **Integrate remaining integral:** $$\int x^2 dx = \frac{x^3}{3}$$ 6. **Substitute back:** $$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$$ 7. **Final answer:** $$\boxed{\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C}$$