1. **State the problem:** We need to evaluate the improper integral $$\int_1^{\infty} \frac{1}{x^2 \sqrt{x}} \, dx$$. 2. **Rewrite the integrand:** Recall that $$\sqrt{x} = x^{1/2}$$, so the integrand becomes $$\frac{1}{x^2 \cdot x^{1/2}} = \frac{1}{x^{2 + \frac{1}{2}}} = \frac{1}{x^{\frac{5}{2}}} = x^{-\frac{5}{2}}$$. 3. **Set up the integral:** $$\int_1^{\infty} x^{-\frac{5}{2}} \, dx$$ 4. **Use the power rule for integration:** For $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$, provided $$n \neq -1$$. Here, $$n = -\frac{5}{2}$$, so $$n+1 = -\frac{5}{2} + 1 = -\frac{3}{2}$$. 5. **Evaluate the definite integral:** $$\int_1^{\infty} x^{-\frac{5}{2}} \, dx = \lim_{t \to \infty} \left[ \frac{x^{-\frac{3}{2}}}{-\frac{3}{2}} \right]_1^t = \lim_{t \to \infty} \left[ -\frac{2}{3} x^{-\frac{3}{2}} \right]_1^t = \lim_{t \to \infty} \left( -\frac{2}{3} t^{-\frac{3}{2}} + \frac{2}{3} \cdot 1^{-\frac{3}{2}} \right)$$ 6. **Calculate the limit:** Since $$t^{-\frac{3}{2}} = \frac{1}{t^{\frac{3}{2}}}$$ and $$t \to \infty$$, this term goes to 0. So the integral evaluates to: $$0 + \frac{2}{3} = \frac{2}{3}$$. **Final answer:** $$\int_1^{\infty} \frac{1}{x^2 \sqrt{x}} \, dx = \frac{2}{3}$$.