1. **State the problem:** Evaluate the integral $$\int -2x \ln(-x) \, dx$$. 2. **Recall the formula for integration by parts:** $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = \ln(-x) \quad \Rightarrow \quad du = \frac{1}{-x} \cdot (-1) \, dx = \frac{1}{x} \, dx$$ $$dv = -2x \, dx \quad \Rightarrow \quad v = \int -2x \, dx = -x^2$$ 3. **Apply integration by parts:** $$\int -2x \ln(-x) \, dx = uv - \int v \, du = (-x^2) \ln(-x) - \int (-x^2) \cdot \frac{1}{x} \, dx$$ 4. **Simplify the integral:** $$\int (-x^2) \cdot \frac{1}{x} \, dx = \int -x \, dx$$ 5. **Evaluate the remaining integral:** $$\int -x \, dx = -\frac{x^2}{2}$$ 6. **Write the full solution:** $$\int -2x \ln(-x) \, dx = -x^2 \ln(-x) - \left(-\frac{x^2}{2}\right) + C = -x^2 \ln(-x) + \frac{x^2}{2} + C$$ **Final answer:** $$\boxed{-x^2 \ln(-x) + \frac{x^2}{2} + C}$$