1. **State the problem:** We want to express the integrand $$\frac{16x^2}{(x-12)(x+4)^2}$$ as a sum of partial fractions and then evaluate the integral $$\int \frac{16x^2}{(x-12)(x+4)^2} \, dx$$. 2. **Set up the partial fraction decomposition:** Since the denominator has a linear factor $(x-12)$ and a repeated linear factor $(x+4)^2$, the decomposition form is: $$\frac{16x^2}{(x-12)(x+4)^2} = \frac{A}{x-12} + \frac{B}{x+4} + \frac{C}{(x+4)^2}$$ 3. **Multiply both sides by the denominator to clear fractions:** $$16x^2 = A(x+4)^2 + B(x-12)(x+4) + C(x-12)$$ 4. **Expand each term:** - $(x+4)^2 = x^2 + 8x + 16$ - $(x-12)(x+4) = x^2 - 8x - 48$ So, $$16x^2 = A(x^2 + 8x + 16) + B(x^2 - 8x - 48) + C(x - 12)$$ 5. **Expand and group terms:** $$16x^2 = (A + B) x^2 + (8A - 8B + C) x + (16A - 48B - 12C)$$ 6. **Equate coefficients of powers of $x$ on both sides:** - Coefficient of $x^2$: $16 = A + B$ - Coefficient of $x$: $0 = 8A - 8B + C$ - Constant term: $0 = 16A - 48B - 12C$ 7. **Solve the system:** From $16 = A + B$, we get $B = 16 - A$. Substitute into the $x$ coefficient equation: $$0 = 8A - 8(16 - A) + C = 8A - 128 + 8A + C = 16A - 128 + C$$ So, $$C = 128 - 16A$$ Substitute $B$ and $C$ into the constant term equation: $$0 = 16A - 48(16 - A) - 12(128 - 16A)$$ $$0 = 16A - 768 + 48A - 1536 + 192A$$ $$0 = (16A + 48A + 192A) - (768 + 1536)$$ $$0 = 256A - 2304$$ Solve for $A$: $$256A = 2304 \implies A = \frac{2304}{256} = 9$$ Then, $$B = 16 - 9 = 7$$ $$C = 128 - 16 \times 9 = 128 - 144 = -16$$ 8. **Write the partial fraction decomposition:** $$\frac{16x^2}{(x-12)(x+4)^2} = \frac{9}{x-12} + \frac{7}{x+4} - \frac{16}{(x+4)^2}$$ 9. **Integrate term-by-term:** $$\int \frac{16x^2}{(x-12)(x+4)^2} \, dx = \int \frac{9}{x-12} \, dx + \int \frac{7}{x+4} \, dx - \int \frac{16}{(x+4)^2} \, dx$$ 10. **Integrate each term:** - $$\int \frac{9}{x-12} \, dx = 9 \ln|x-12| + C_1$$ - $$\int \frac{7}{x+4} \, dx = 7 \ln|x+4| + C_2$$ - $$\int \frac{16}{(x+4)^2} \, dx = 16 \int (x+4)^{-2} \, dx = 16 \left(-\frac{1}{x+4}\right) + C_3 = -\frac{16}{x+4} + C_3$$ 11. **Combine the results:** $$\int \frac{16x^2}{(x-12)(x+4)^2} \, dx = 9 \ln|x-12| + 7 \ln|x+4| - \frac{16}{x+4} + C$$ **Final answer:** $$\boxed{9 \ln|x-12| + 7 \ln|x+4| - \frac{16}{x+4} + C}$$