1. The problem is to evaluate the integral $$\int \frac{3x^2}{\sqrt[3]{x-1}} \, dx$$. 2. Rewrite the integral by expressing the cube root in the denominator as a power: $$\sqrt[3]{x-1} = (x-1)^{\frac{1}{3}}$$, so the integral becomes $$\int 3x^2 (x-1)^{-\frac{1}{3}} \, dx$$. 3. Use substitution to simplify the integral. Let $$u = x - 1$$, then $$du = dx$$ and $$x = u + 1$$. 4. Substitute into the integral: $$\int 3(u+1)^2 u^{-\frac{1}{3}} \, du$$. 5. Expand $$(u+1)^2 = u^2 + 2u + 1$$, so the integral becomes: $$\int 3(u^2 + 2u + 1) u^{-\frac{1}{3}} \, du = \int 3(u^{2 - \frac{1}{3}} + 2u^{1 - \frac{1}{3}} + u^{0 - \frac{1}{3}}) \, du$$. 6. Simplify the exponents: $$2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}$$, $$1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$, $$0 - \frac{1}{3} = -\frac{1}{3}$$. 7. The integral is now: $$\int 3(u^{\frac{5}{3}} + 2u^{\frac{2}{3}} + u^{-\frac{1}{3}}) \, du = 3 \int u^{\frac{5}{3}} \, du + 6 \int u^{\frac{2}{3}} \, du + 3 \int u^{-\frac{1}{3}} \, du$$. 8. Integrate each term using the power rule $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$: - $$3 \int u^{\frac{5}{3}} \, du = 3 \cdot \frac{u^{\frac{5}{3} + 1}}{\frac{5}{3} + 1} = 3 \cdot \frac{u^{\frac{8}{3}}}{\frac{8}{3}} = 3 \cdot \frac{3}{8} u^{\frac{8}{3}} = \frac{9}{8} u^{\frac{8}{3}}$$ - $$6 \int u^{\frac{2}{3}} \, du = 6 \cdot \frac{u^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} = 6 \cdot \frac{u^{\frac{5}{3}}}{\frac{5}{3}} = 6 \cdot \frac{3}{5} u^{\frac{5}{3}} = \frac{18}{5} u^{\frac{5}{3}}$$ - $$3 \int u^{-\frac{1}{3}} \, du = 3 \cdot \frac{u^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} = 3 \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} = 3 \cdot \frac{3}{2} u^{\frac{2}{3}} = \frac{9}{2} u^{\frac{2}{3}}$$ 9. Combine the results: $$\frac{9}{8} u^{\frac{8}{3}} + \frac{18}{5} u^{\frac{5}{3}} + \frac{9}{2} u^{\frac{2}{3}} + C$$ 10. Substitute back $$u = x - 1$$: $$\boxed{\frac{9}{8} (x-1)^{\frac{8}{3}} + \frac{18}{5} (x-1)^{\frac{5}{3}} + \frac{9}{2} (x-1)^{\frac{2}{3}} + C}$$ This is the evaluated integral.