1. **State the problem:** Evaluate the integral $$\int \sqrt{x^3 + 2} \cdot x^2 \, dx.$$\n\n2. **Identify substitution:** Let $$u = x^3 + 2.$$ Then, differentiate both sides with respect to $$x$$:\n$$\frac{du}{dx} = 3x^2 \implies du = 3x^2 \, dx.$$\n\n3. **Rewrite the integral:** We have $$x^2 \, dx = \frac{du}{3}.$$ Substitute into the integral:\n$$\int \sqrt{u} \cdot \frac{du}{3} = \frac{1}{3} \int u^{1/2} \, du.$$\n\n4. **Integrate:** Use the power rule for integration:\n$$\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C,$$ where $$n = \frac{1}{2}.$$\nSo,\n$$\frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9} u^{3/2} + C.$$\n\n5. **Back-substitute:** Replace $$u$$ with $$x^3 + 2$$:\n$$\frac{2}{9} (x^3 + 2)^{3/2} + C.$$\n\n**Final answer:** $$\int \sqrt{x^3 + 2} \cdot x^2 \, dx = \frac{2}{9} (x^3 + 2)^{3/2} + C.$$