1. **State the problem:** Find the integral $$\int \frac{x^2}{\sqrt{x-1}} \, dx$$. 2. **Rewrite the integral:** Let us use substitution to simplify the integral. Set $$u = x - 1$$, so $$x = u + 1$$ and $$dx = du$$. 3. **Rewrite the integral in terms of $u$:** $$\int \frac{(u+1)^2}{\sqrt{u}} \, du = \int \frac{u^2 + 2u + 1}{u^{1/2}} \, du = \int (u^{3/2} + 2u^{1/2} + u^{-1/2}) \, du$$. 4. **Integrate each term separately:** - $$\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$$ - $$\int 2u^{1/2} \, du = 2 \cdot \frac{u^{3/2}}{3/2} = \frac{4}{3} u^{3/2}$$ - $$\int u^{-1/2} \, du = 2 u^{1/2}$$ 5. **Combine the results:** $$\int \frac{x^2}{\sqrt{x-1}} \, dx = \frac{2}{5} u^{5/2} + \frac{4}{3} u^{3/2} + 2 u^{1/2} + C$$ 6. **Substitute back $u = x - 1$:** $$= \frac{2}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + 2 (x-1)^{1/2} + C$$ **Final answer:** $$\int \frac{x^2}{\sqrt{x-1}} \, dx = \frac{2}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + 2 (x-1)^{1/2} + C$$