1. **State the problem:** We need to evaluate the definite integral $$\int_{-2}^1 x \sqrt{3+x} \, dx$$. 2. **Substitution:** Let $$u = 3 + x$$, then $$du = dx$$ and when $$x = -2$$, $$u = 1$$; when $$x = 1$$, $$u = 4$$. 3. Express $$x$$ in terms of $$u$$: $$x = u - 3$$. 4. Rewrite the integral in terms of $$u$$: $$\int_{u=1}^{4} (u - 3) \sqrt{u} \, du = \int_1^4 (u - 3) u^{1/2} \, du = \int_1^4 (u^{3/2} - 3u^{1/2}) \, du$$. 5. **Integrate term-by-term:** $$\int u^{3/2} \, du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}$$ $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}$$ 6. So the integral becomes: $$\left[ \frac{2}{5} u^{5/2} - 3 \cdot \frac{2}{3} u^{3/2} \right]_1^4 = \left[ \frac{2}{5} u^{5/2} - 2 u^{3/2} \right]_1^4$$. 7. **Evaluate at the bounds:** At $$u=4$$: $$\frac{2}{5} \cdot 4^{5/2} - 2 \cdot 4^{3/2} = \frac{2}{5} \cdot (4^{2} \cdot 4^{1/2}) - 2 \cdot (4^{1} \cdot 4^{1/2}) = \frac{2}{5} \cdot (16 \cdot 2) - 2 \cdot (4 \cdot 2) = \frac{2}{5} \cdot 32 - 2 \cdot 8 = \frac{64}{5} - 16$$ At $$u=1$$: $$\frac{2}{5} \cdot 1^{5/2} - 2 \cdot 1^{3/2} = \frac{2}{5} - 2 = \frac{2}{5} - \frac{10}{5} = -\frac{8}{5}$$ 8. **Subtract:** $$\left( \frac{64}{5} - 16 \right) - \left( -\frac{8}{5} \right) = \frac{64}{5} - 16 + \frac{8}{5} = \frac{64 + 8}{5} - 16 = \frac{72}{5} - 16 = \frac{72}{5} - \frac{80}{5} = -\frac{8}{5}$$ **Final answer:** $$\boxed{-\frac{8}{5}}$$