1. **State the problem:** We need to evaluate the integral $$\int x e^{-2x} \, dx$$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = x \implies du = dx$$ $$dv = e^{-2x} dx \implies v = \int e^{-2x} dx = -\frac{1}{2} e^{-2x}$$ 3. **Apply integration by parts:** $$\int x e^{-2x} dx = u v - \int v \, du = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) dx$$ 4. **Simplify the expression:** $$= -\frac{x}{2} e^{-2x} + \frac{1}{2} \int e^{-2x} dx$$ 5. **Integrate remaining integral:** $$\int e^{-2x} dx = -\frac{1}{2} e^{-2x}$$ 6. **Substitute back:** $$= -\frac{x}{2} e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) + C = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C$$ 7. **Final answer:** $$\int x e^{-2x} dx = -\frac{e^{-2x}}{2} \left(x + \frac{1}{2}\right) + C$$