1. **State the problem:** Evaluate the integral $$\int \frac{x^3}{\sqrt{2-3x^4}} \, dx$$. 2. **Identify substitution:** Let $$u = 2 - 3x^4$$. Then, differentiate: $$du = -12x^3 \, dx$$. 3. **Rewrite differential:** Solve for $$x^3 \, dx$$: $$x^3 \, dx = -\frac{1}{12} du$$. 4. **Substitute into integral:** Replace $$x^3 \, dx$$ and the expression under the root: $$\int \frac{x^3}{\sqrt{2-3x^4}} \, dx = \int \frac{-\frac{1}{12} du}{\sqrt{u}} = -\frac{1}{12} \int u^{-\frac{1}{2}} \, du$$. 5. **Integrate:** Use the power rule for integrals: $$\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. Here, $$n = -\frac{1}{2}$$, so $$\int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} + C$$. 6. **Apply result:** $$-\frac{1}{12} \times 2u^{\frac{1}{2}} + C = -\frac{1}{6} \sqrt{u} + C$$. 7. **Back-substitute:** Replace $$u$$ with original expression: $$-\frac{1}{6} \sqrt{2 - 3x^4} + C$$. **Final answer:** $$\boxed{-\frac{1}{6} \sqrt{2 - 3x^4} + C}$$