1. The problem asks to evaluate the indefinite integral $$\int xe^{3x} \, dx$$. 2. To solve this, we use integration by parts. Recall the formula: $$\int u \, dv = uv - \int v \, du$$. 3. Let $$u = x$$, which implies $$du = dx$$. 4. Let $$dv = e^{3x} dx$$, which implies $$v = \frac{1}{3} e^{3x}$$ because $$\int e^{ax} dx = \frac{1}{a} e^{ax}$$. 5. Using the integration by parts formula: $$\int xe^{3x} dx = x \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cdot 1 \, dx = \frac{x}{3} e^{3x} - \frac{1}{3} \int e^{3x} dx$$. 6. Now evaluate $$\int e^{3x} dx = \frac{1}{3} e^{3x}$$. 7. Substitute back: $$\int xe^{3x} dx = \frac{x}{3} e^{3x} - \frac{1}{3} \cdot \frac{1}{3} e^{3x} + C = \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x} + C$$. 8. Factor out $$e^{3x}$$: $$\int xe^{3x} dx = e^{3x} \left( \frac{x}{3} - \frac{1}{9} \right) + C$$. **Answer:** $$\boxed{e^{3x} \left( \frac{x}{3} - \frac{1}{9} \right) + C}$$