1. Planteamos el problema: calcular la integral indefinida $$\int \frac{x^2}{\sqrt{2x+1}} \, dx$$. 2. Usamos el cambio de variable para simplificar la raíz. Sea $$u = 2x + 1$$, entonces $$du = 2 dx$$ o $$dx = \frac{du}{2}$$. 3. Expresamos $$x$$ en términos de $$u$$: $$x = \frac{u - 1}{2}$$. 4. Reescribimos el integrando en función de $$u$$: $$\frac{x^2}{\sqrt{2x+1}} dx = \frac{\left(\frac{u-1}{2}\right)^2}{\sqrt{u}} \cdot \frac{du}{2} = \frac{(u-1)^2}{4} \cdot \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{(u-1)^2}{8 \sqrt{u}} du$$. 5. Expandimos el numerador: $$(u-1)^2 = u^2 - 2u + 1$$. 6. La integral queda: $$\int \frac{u^2 - 2u + 1}{8 \sqrt{u}} du = \frac{1}{8} \int \frac{u^2 - 2u + 1}{u^{1/2}} du = \frac{1}{8} \int \left(u^{2 - \frac{1}{2}} - 2u^{1 - \frac{1}{2}} + u^{0 - \frac{1}{2}}\right) du = \frac{1}{8} \int \left(u^{\frac{3}{2}} - 2u^{\frac{1}{2}} + u^{-\frac{1}{2}}\right) du$$. 7. Integramos término a término: $$\int u^{\frac{3}{2}} du = \frac{u^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} u^{\frac{5}{2}}$$ $$\int u^{\frac{1}{2}} du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} u^{\frac{3}{2}}$$ $$\int u^{-\frac{1}{2}} du = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2 u^{\frac{1}{2}}$$. 8. Sustituimos en la integral: $$\frac{1}{8} \left( \frac{2}{5} u^{\frac{5}{2}} - 2 \cdot \frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} \right) + C = \frac{1}{8} \left( \frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} \right) + C$$. 9. Simplificamos los coeficientes: $$= \frac{1}{8} \cdot \frac{2}{5} u^{\frac{5}{2}} - \frac{1}{8} \cdot \frac{4}{3} u^{\frac{3}{2}} + \frac{1}{8} \cdot 2 u^{\frac{1}{2}} + C = \frac{1}{20} u^{\frac{5}{2}} - \frac{1}{6} u^{\frac{3}{2}} + \frac{1}{4} u^{\frac{1}{2}} + C$$. 10. Finalmente, regresamos a la variable original $$x$$: $$u = 2x + 1$$, por lo que $$\int \frac{x^2}{\sqrt{2x+1}} dx = \frac{1}{20} (2x+1)^{\frac{5}{2}} - \frac{1}{6} (2x+1)^{\frac{3}{2}} + \frac{1}{4} (2x+1)^{\frac{1}{2}} + C$$.