1. **State the problem:** We want to evaluate the integral $$\int (2x^2 + 1)^{\frac{1}{3}} x^3 \, dx.$$\n\n2. **Use u-substitution:** Let $$u = 2x^2 + 1.$$ Then, differentiate both sides with respect to $$x$$:\n$$\frac{du}{dx} = 4x \implies du = 4x \, dx.$$\n\n3. **Rewrite the integral in terms of $$u$$:** We have $$x^3 dx$$ in the integral, but from $$du = 4x dx$$ we get $$x dx = \frac{du}{4}$$. To express $$x^3 dx$$, write it as $$x^2 \cdot x dx$$. Since $$u = 2x^2 + 1$$, we can solve for $$x^2$$:\n$$x^2 = \frac{u - 1}{2}.$$\n\n4. Substitute into the integral:\n$$\int (2x^2 + 1)^{\frac{1}{3}} x^3 \, dx = \int u^{\frac{1}{3}} \cdot x^2 \cdot x dx = \int u^{\frac{1}{3}} \cdot \frac{u - 1}{2} \cdot \frac{du}{4} = \int u^{\frac{1}{3}} \cdot \frac{u - 1}{2} \cdot \frac{du}{4}.$$\n\n5. Simplify constants:\n$$\int u^{\frac{1}{3}} \cdot \frac{u - 1}{2} \cdot \frac{du}{4} = \int \frac{u^{\frac{1}{3}} (u - 1)}{8} \, du = \frac{1}{8} \int u^{\frac{1}{3}} (u - 1) \, du.$$\n\n6. Expand the integrand:\n$$u^{\frac{1}{3}} (u - 1) = u^{\frac{1}{3}} \cdot u - u^{\frac{1}{3}} = u^{\frac{4}{3}} - u^{\frac{1}{3}}.$$\n\n7. The integral becomes:\n$$\frac{1}{8} \int (u^{\frac{4}{3}} - u^{\frac{1}{3}}) \, du = \frac{1}{8} \left( \int u^{\frac{4}{3}} \, du - \int u^{\frac{1}{3}} \, du \right).$$\n\n8. Integrate each term using the power rule $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$:\n$$\int u^{\frac{4}{3}} du = \frac{u^{\frac{4}{3} + 1}}{\frac{4}{3} + 1} = \frac{u^{\frac{7}{3}}}{\frac{7}{3}} = \frac{3}{7} u^{\frac{7}{3}},$$\n$$\int u^{\frac{1}{3}} du = \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{u^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4} u^{\frac{4}{3}}.$$\n\n9. Substitute back into the integral:\n$$\frac{1}{8} \left( \frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}} \right) + C = \frac{3}{8} \left( \frac{1}{7} u^{\frac{7}{3}} - \frac{1}{4} u^{\frac{4}{3}} \right) + C = \frac{3}{56} u^{\frac{7}{3}} - \frac{3}{32} u^{\frac{4}{3}} + C.$$\n\n10. Finally, substitute back $$u = 2x^2 + 1$$ to get the answer:\n$$\boxed{\frac{3}{56} (2x^2 + 1)^{\frac{7}{3}} - \frac{3}{32} (2x^2 + 1)^{\frac{4}{3}} + C}.$$