1. **State the problem:** Multiply and divide the rational expressions: $$\frac{4x^3+4x^2}{2x} \div \frac{x^2-1}{x^2+2x-3} \cdot \frac{x+3}{x^2+3x}$$ 2. **Rewrite division as multiplication by reciprocal:** $$\frac{4x^3+4x^2}{2x} \times \frac{x^2+2x-3}{x^2-1} \times \frac{x+3}{x^2+3x}$$ 3. **Factor all polynomials:** - Numerator and denominator of first fraction: $$4x^3+4x^2 = 4x^2(x+1)$$ $$2x = 2x$$ - Factor second fraction numerator and denominator: $$x^2+2x-3 = (x+3)(x-1)$$ $$x^2-1 = (x-1)(x+1)$$ - Factor denominator of third fraction: $$x^2+3x = x(x+3)$$ 4. **Rewrite expression with factors:** $$\frac{4x^2(x+1)}{2x} \times \frac{(x+3)(x-1)}{(x-1)(x+1)} \times \frac{x+3}{x(x+3)}$$ 5. **Cancel common factors step-by-step:** - Cancel $x$ in first fraction: $$\frac{4x^2(x+1)}{\cancel{2x}} = \frac{4x^2(x+1)}{2\cancel{x}}$$ - Simplify $\frac{4x^2}{2} = 2x^2$: $$2x^2(x+1)$$ - Cancel $(x-1)$ in second fraction numerator and denominator: $$\frac{(x+3)\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} = \frac{x+3}{x+1}$$ - Cancel $(x+3)$ in third fraction numerator and denominator: $$\frac{\cancel{x+3}}{x\cancel{(x+3)}} = \frac{1}{x}$$ 6. **Multiply the simplified expressions:** $$2x^2(x+1) \times \frac{x+3}{x+1} \times \frac{1}{x}$$ 7. **Cancel $(x+1)$:** $$2x^2\cancel{(x+1)} \times \frac{x+3}{\cancel{(x+1)}} \times \frac{1}{x} = 2x^2 \times (x+3) \times \frac{1}{x}$$ 8. **Simplify $\frac{2x^2}{x} = 2x$:** $$2x(x+3)$$ 9. **Final simplified expression:** $$\boxed{2x(x+3)}$$