1. **State the problem:** We need to evaluate the integral $$\int x^2 \sqrt{x-1} \, dx$$ using the method of substitution. 2. **Choose a substitution:** Let $$u = x - 1$$. Then, $$du = dx$$ and $$x = u + 1$$. 3. **Rewrite the integral in terms of $$u$$:** $$\int x^2 \sqrt{x-1} \, dx = \int (u+1)^2 \sqrt{u} \, du$$ 4. **Expand and simplify:** $$(u+1)^2 = u^2 + 2u + 1$$ So the integral becomes: $$\int (u^2 + 2u + 1) u^{1/2} \, du = \int (u^{2} u^{1/2} + 2u u^{1/2} + 1 \cdot u^{1/2}) \, du$$ $$= \int (u^{5/2} + 2u^{3/2} + u^{1/2}) \, du$$ 5. **Integrate term-by-term:** Use the power rule $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. $$\int u^{5/2} du = \frac{u^{7/2}}{7/2} = \frac{2}{7} u^{7/2}$$ $$\int 2u^{3/2} du = 2 \cdot \frac{u^{5/2}}{5/2} = 2 \cdot \frac{2}{5} u^{5/2} = \frac{4}{5} u^{5/2}$$ $$\int u^{1/2} du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$$ 6. **Combine the results:** $$\int x^2 \sqrt{x-1} \, dx = \frac{2}{7} u^{7/2} + \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$$ 7. **Substitute back $$u = x - 1$$:** $$= \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C$$ **Final answer:** $$\int x^2 \sqrt{x-1} \, dx = \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C$$