1. **State the problem:** We need to evaluate the integral $$\int \frac{x^2}{1-2x^3} \, dx.$$\n\n2. **Identify a substitution:** Notice the denominator is $1-2x^3$. Its derivative is $-6x^2$, which is similar to the numerator $x^2$. This suggests using substitution.\n\n3. **Substitution:** Let $$u = 1 - 2x^3.$$ Then, differentiate both sides: $$\frac{du}{dx} = -6x^2 \implies du = -6x^2 \, dx.$$\n\n4. **Rewrite the integral:** Solve for $x^2 \, dx$: $$x^2 \, dx = -\frac{1}{6} du.$$ Substitute into the integral: $$\int \frac{x^2}{1-2x^3} \, dx = \int \frac{1}{u} \left(-\frac{1}{6} du\right) = -\frac{1}{6} \int \frac{1}{u} \, du.$$\n\n5. **Integrate:** The integral of $\frac{1}{u}$ is $\ln|u|$, so: $$-\frac{1}{6} \int \frac{1}{u} \, du = -\frac{1}{6} \ln|u| + C.$$\n\n6. **Back-substitute:** Replace $u$ with $1 - 2x^3$: $$-\frac{1}{6} \ln|1 - 2x^3| + C.$$\n\n**Final answer:** $$\boxed{-\frac{1}{6} \ln|1 - 2x^3| + C}.$$