1. **Stating the problem:** Calculate the integral $$\int x \sqrt{x^2+3} \, dx = \int x (x^2+3)^{1/2} \, dx$$. 2. **Formula and substitution:** Use substitution method. Let $$u = x^2 + 3$$. Then, $$\frac{du}{dx} = 2x$$, so $$du = 2x \, dx$$. 3. **Rewrite the integral:** From $$du = 2x \, dx$$, we get $$x \, dx = \frac{du}{2}$$. Substitute into the integral: $$\int x (x^2+3)^{1/2} \, dx = \int (u)^{1/2} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/2} \, du$$. 4. **Integrate:** Use the power rule for integration: $$\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. Here, $$n = \frac{1}{2}$$, so $$\frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$. 5. **Back-substitute:** Replace $$u$$ with $$x^2 + 3$$: $$\frac{1}{3} (x^2 + 3)^{3/2} + C$$. **Final answer:** $$\int x \sqrt{x^2+3} \, dx = \frac{1}{3} (x^2 + 3)^{3/2} + C$$.