1. **State the problem:** Find the indefinite integral of the function $3x \cos(3x)$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ Choose $u = 3x$ and $dv = \cos(3x) \, dx$. 3. **Calculate derivatives and integrals:** $$du = 3 \, dx$$ $$v = \int \cos(3x) \, dx = \frac{\sin(3x)}{3}$$ 4. **Apply integration by parts:** $$\int 3x \cos(3x) \, dx = uv - \int v \, du = 3x \cdot \frac{\sin(3x)}{3} - \int \frac{\sin(3x)}{3} \cdot 3 \, dx$$ 5. **Simplify the expression:** $$= x \sin(3x) - \int \sin(3x) \, dx$$ 6. **Integrate remaining integral:** $$\int \sin(3x) \, dx = -\frac{\cos(3x)}{3}$$ 7. **Substitute back:** $$x \sin(3x) - \left(-\frac{\cos(3x)}{3}\right) + C = x \sin(3x) + \frac{\cos(3x)}{3} + C$$ **Final answer:** $$\int 3x \cos(3x) \, dx = x \sin(3x) + \frac{\cos(3x)}{3} + C$$