1. We are asked to evaluate the definite integral $$\int_0^1 \frac{x^5}{x^3 + 2} \, dx.$$\n\n2. The integral involves a rational function where the numerator and denominator are polynomials. A useful approach is to simplify the integrand by algebraic manipulation.\n\n3. Notice that $$x^5 = x^2 \cdot x^3,$$ so we can rewrite the integrand as $$\frac{x^5}{x^3 + 2} = \frac{x^2 \cdot x^3}{x^3 + 2}.$$\n\n4. Rewrite the numerator to express the integrand as $$\frac{x^2 (x^3 + 2 - 2)}{x^3 + 2} = x^2 - \frac{2x^2}{x^3 + 2}.$$\n\n5. Therefore, the integral becomes $$\int_0^1 \left(x^2 - \frac{2x^2}{x^3 + 2}\right) dx = \int_0^1 x^2 \, dx - 2 \int_0^1 \frac{x^2}{x^3 + 2} \, dx.$$\n\n6. The first integral is straightforward: $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}.$$\n\n7. For the second integral, use substitution: let $$u = x^3 + 2,$$ then $$du = 3x^2 dx,$$ so $$x^2 dx = \frac{du}{3}.$$\n\n8. Change the limits for $$u$$: when $$x=0,$$ $$u=0^3 + 2 = 2,$$ and when $$x=1,$$ $$u=1^3 + 2 = 3.$$\n\n9. The integral becomes $$\int_0^1 \frac{x^2}{x^3 + 2} dx = \int_2^3 \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int_2^3 \frac{1}{u} du = \frac{1}{3} [\ln|u|]_2^3 = \frac{1}{3} (\ln 3 - \ln 2).$$\n\n10. Substitute back into the expression: $$\int_0^1 \frac{x^5}{x^3 + 2} dx = \frac{1}{3} - 2 \cdot \frac{1}{3} (\ln 3 - \ln 2) = \frac{1}{3} - \frac{2}{3} \ln \frac{3}{2}.$$\n\n11. Final answer: $$\boxed{\frac{1}{3} - \frac{2}{3} \ln \frac{3}{2}}.$$