1. **State the problem:** Evaluate the integral $$\int \frac{4}{x^2 \sqrt{x}} \, dx$$. 2. **Rewrite the integrand:** Recall that $$\sqrt{x} = x^{\frac{1}{2}}$$, so the integrand becomes $$\frac{4}{x^2 \cdot x^{\frac{1}{2}}} = \frac{4}{x^{2 + \frac{1}{2}}} = \frac{4}{x^{\frac{5}{2}}}$$. 3. **Express the integrand with negative exponents:** $$4x^{-\frac{5}{2}}$$. 4. **Use the power rule for integration:** For $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$, provided $$n \neq -1$$. 5. **Apply the rule:** Here, $$n = -\frac{5}{2}$$, so $$\int 4x^{-\frac{5}{2}} \, dx = 4 \int x^{-\frac{5}{2}} \, dx = 4 \cdot \frac{x^{-\frac{5}{2} + 1}}{-\frac{5}{2} + 1} + C = 4 \cdot \frac{x^{-\frac{3}{2}}}{-\frac{3}{2}} + C$$. 6. **Simplify the denominator:** $$-\frac{3}{2} = -1.5$$, so $$4 \cdot \frac{x^{-\frac{3}{2}}}{-\frac{3}{2}} = 4 \cdot \left(-\frac{2}{3}\right) x^{-\frac{3}{2}} = -\frac{8}{3} x^{-\frac{3}{2}} + C$$. 7. **Rewrite the answer in radical form:** Since $$x^{-\frac{3}{2}} = \frac{1}{x^{\frac{3}{2}}} = \frac{1}{x \sqrt{x}}$$, final answer is $$-\frac{8}{3} \cdot \frac{1}{x \sqrt{x}} + C = -\frac{8}{3x \sqrt{x}} + C$$. **Final answer:** $$\boxed{-\frac{8}{3x \sqrt{x}} + C}$$