1. **State the problem:** We need to evaluate the integral $$\int x \sqrt{x^2 + 9} \, dx$$. 2. **Identify the method:** This integral suggests using substitution because the integrand contains a function and its derivative. 3. **Substitution:** Let $$u = x^2 + 9$$. Then, $$\frac{du}{dx} = 2x$$, so $$du = 2x \, dx$$. 4. **Rewrite the integral:** From $$du = 2x \, dx$$, we get $$x \, dx = \frac{du}{2}$$. Substitute into the integral: $$\int x \sqrt{x^2 + 9} \, dx = \int \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/2} \, du$$. 5. **Integrate:** Use the power rule for integration: $$\frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{\frac{3}{2}} = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} + C$$. 6. **Back-substitute:** Replace $$u$$ with $$x^2 + 9$$: $$\frac{1}{3} (x^2 + 9)^{3/2} + C$$. **Final answer:** $$\int x \sqrt{x^2 + 9} \, dx = \frac{1}{3} (x^2 + 9)^{3/2} + C$$.