1. We are asked to evaluate the definite integral $$\int_0^1 \frac{x^2}{\sqrt{x^3 + 1}} \, dx$$. 2. To solve this integral, we use substitution. Let $$u = x^3 + 1$$. Then, $$\frac{du}{dx} = 3x^2$$, so $$du = 3x^2 dx$$. 3. Rearranging for $$x^2 dx$$, we get $$x^2 dx = \frac{du}{3}$$. 4. Change the limits of integration: when $$x=0$$, $$u=0^3 + 1 = 1$$; when $$x=1$$, $$u=1^3 + 1 = 2$$. 5. Substitute into the integral: $$\int_0^1 \frac{x^2}{\sqrt{x^3 + 1}} dx = \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{du}{3} = \frac{1}{3} \int_1^2 u^{-\frac{1}{2}} du$$. 6. Integrate: $$\int u^{-\frac{1}{2}} du = 2 u^{\frac{1}{2}} + C$$. 7. So, $$\frac{1}{3} \int_1^2 u^{-\frac{1}{2}} du = \frac{1}{3} \cdot 2 u^{\frac{1}{2}} \Big|_1^2 = \frac{2}{3} (\sqrt{2} - 1)$$. 8. Therefore, the value of the integral is: $$\boxed{\frac{2}{3} (\sqrt{2} - 1)}$$.