1. **State the problem:** We need to find the integral $$\int \frac{x^2}{(x-2)^{10}} \, dx$$. 2. **Rewrite the integrand:** Let’s use substitution to simplify the integral. Set $$u = x - 2$$, so $$x = u + 2$$ and $$dx = du$$. 3. **Express the integral in terms of $u$:** $$\int \frac{(u+2)^2}{u^{10}} \, du = \int \frac{u^2 + 4u + 4}{u^{10}} \, du = \int \left(u^{2-10} + 4u^{1-10} + 4u^{-10}\right) \, du = \int \left(u^{-8} + 4u^{-9} + 4u^{-10}\right) \, du$$ 4. **Integrate term-by-term:** - $$\int u^{-8} \, du = \frac{u^{-7}}{-7} + C_1 = -\frac{1}{7} u^{-7} + C_1$$ - $$\int 4u^{-9} \, du = 4 \cdot \frac{u^{-8}}{-8} + C_2 = -\frac{1}{2} u^{-8} + C_2$$ - $$\int 4u^{-10} \, du = 4 \cdot \frac{u^{-9}}{-9} + C_3 = -\frac{4}{9} u^{-9} + C_3$$ 5. **Combine the results:** $$-\frac{1}{7} u^{-7} - \frac{1}{2} u^{-8} - \frac{4}{9} u^{-9} + C$$ 6. **Substitute back $u = x - 2$:** $$-\frac{1}{7 (x-2)^7} - \frac{1}{2 (x-2)^8} - \frac{4}{9 (x-2)^9} + C$$ **Final answer:** $$\int \frac{x^2}{(x-2)^{10}} \, dx = -\frac{1}{7 (x-2)^7} - \frac{1}{2 (x-2)^8} - \frac{4}{9 (x-2)^9} + C$$