1. **State the problem:** We need to find the integral $$\int \frac{x^2}{3} e^{3x} \, dx$$. 2. **Rewrite the integral:** Factor out the constant $\frac{1}{3}$: $$\int \frac{x^2}{3} e^{3x} \, dx = \frac{1}{3} \int x^2 e^{3x} \, dx$$ 3. **Use integration by parts:** Recall the formula: $$\int u \, dv = uv - \int v \, du$$ 4. **Choose parts:** Let $u = x^2$ so $du = 2x \, dx$, $dv = e^{3x} \, dx$ so $v = \frac{1}{3} e^{3x}$. 5. **Apply integration by parts:** $$\int x^2 e^{3x} \, dx = x^2 \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cdot 2x \, dx = \frac{x^2}{3} e^{3x} - \frac{2}{3} \int x e^{3x} \, dx$$ 6. **Integrate $\int x e^{3x} \, dx$ by parts again:** Let $u = x$, $du = dx$, $dv = e^{3x} dx$, $v = \frac{1}{3} e^{3x}$. 7. **Apply integration by parts again:** $$\int x e^{3x} \, dx = x \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \, dx = \frac{x}{3} e^{3x} - \frac{1}{3} \int e^{3x} \, dx$$ 8. **Integrate $\int e^{3x} \, dx$:** $$\int e^{3x} \, dx = \frac{1}{3} e^{3x} + C$$ 9. **Substitute back:** $$\int x e^{3x} \, dx = \frac{x}{3} e^{3x} - \frac{1}{3} \cdot \frac{1}{3} e^{3x} = \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x}$$ 10. **Substitute into step 5:** $$\int x^2 e^{3x} \, dx = \frac{x^2}{3} e^{3x} - \frac{2}{3} \left( \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x} \right) = \frac{x^2}{3} e^{3x} - \frac{2x}{9} e^{3x} + \frac{2}{27} e^{3x}$$ 11. **Recall the original integral:** $$\int \frac{x^2}{3} e^{3x} \, dx = \frac{1}{3} \int x^2 e^{3x} \, dx = \frac{1}{3} \left( \frac{x^2}{3} e^{3x} - \frac{2x}{9} e^{3x} + \frac{2}{27} e^{3x} \right)$$ 12. **Simplify:** $$= \frac{x^2}{9} e^{3x} - \frac{2x}{27} e^{3x} + \frac{2}{81} e^{3x} + C$$ **Final answer:** $$\boxed{\int \frac{x^2}{3} e^{3x} \, dx = e^{3x} \left( \frac{x^2}{9} - \frac{2x}{27} + \frac{2}{81} \right) + C}$$