1. **State the problem:** We need to evaluate the integral $$\int \frac{x^2}{\sqrt{2x+1}} \, dx$$. 2. **Use substitution:** Let $$u = 2x + 1$$, then $$du = 2 \, dx$$ or $$dx = \frac{du}{2}$$. 3. **Express $$x$$ in terms of $$u$$:** From $$u = 2x + 1$$, we get $$x = \frac{u - 1}{2}$$. 4. **Rewrite the integral:** Substitute $$x$$ and $$dx$$: $$\int \frac{x^2}{\sqrt{2x+1}} \, dx = \int \frac{\left(\frac{u - 1}{2}\right)^2}{\sqrt{u}} \cdot \frac{du}{2} = \int \frac{(u - 1)^2}{4} \cdot \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \int \frac{(u - 1)^2}{8 \sqrt{u}} \, du$$ 5. **Simplify the integrand:** $$\frac{(u - 1)^2}{8 \sqrt{u}} = \frac{u^2 - 2u + 1}{8 u^{1/2}} = \frac{u^2}{8 u^{1/2}} - \frac{2u}{8 u^{1/2}} + \frac{1}{8 u^{1/2}} = \frac{u^{3/2}}{8} - \frac{2 u^{1/2}}{8} + \frac{1}{8} u^{-1/2}$$ 6. **Rewrite integral as sum:** $$\int \left( \frac{u^{3/2}}{8} - \frac{2 u^{1/2}}{8} + \frac{1}{8} u^{-1/2} \right) du = \frac{1}{8} \int u^{3/2} du - \frac{2}{8} \int u^{1/2} du + \frac{1}{8} \int u^{-1/2} du$$ 7. **Integrate each term:** - $$\int u^{3/2} du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$$ - $$\int u^{1/2} du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$$ - $$\int u^{-1/2} du = 2 u^{1/2}$$ 8. **Substitute back:** $$\frac{1}{8} \cdot \frac{2}{5} u^{5/2} - \frac{2}{8} \cdot \frac{2}{3} u^{3/2} + \frac{1}{8} \cdot 2 u^{1/2} + C = \frac{1}{20} u^{5/2} - \frac{1}{6} u^{3/2} + \frac{1}{4} u^{1/2} + C$$ 9. **Replace $$u$$ with $$2x + 1$$:** $$\boxed{\frac{(2x+1)^{5/2}}{20} - \frac{(2x+1)^{3/2}}{6} + \frac{(2x+1)^{1/2}}{4} + C}$$