1. **Stating the problem:** We need to compute the integral $$\int \frac{16x^2}{(x-18)(x+6)^2} \, dx.$$ 2. **Set up partial fractions:** Because the denominator factors into a linear term and a squared linear term, express as $$\frac{16x^2}{(x-18)(x+6)^2} = \frac{A}{x-18} + \frac{B}{x+6} + \frac{C}{(x+6)^2}.$$ 3. **Multiply both sides by the denominator:** $$16x^2 = A(x+6)^2 + B(x-18)(x+6) + C(x-18).$$ 4. **Expand terms:** $$A(x^2 + 12x + 36) + B(x^2 - 12x - 108) + C(x - 18).$$ 5. **Group by powers:** $$x^2: A + B$$ $$x: 12A - 12B + C$$ $$constant: 36A - 108B - 18C$$ 6. **Match coefficients with left side $16x^2 + 0x + 0$:** $$A + B = 16$$ $$12A - 12B + C = 0$$ $$36A - 108B -18C = 0.$$ 7. **Solve system:** From the first, $B = 16 - A$. Substitute into second: $$12A - 12(16 - A) + C = 0 \Rightarrow 12A - 192 + 12A + C = 0 \Rightarrow 24A + C = 192,$$ so $C = 192 - 24A$. Substitute $B$ and $C$ into third: $$36A - 108(16 - A) - 18(192 - 24A) = 0,$$ $$36A - 1728 + 108A - 3456 + 432A = 0,$$ sum coefficients for $A$: $$36A + 108A + 432A = 576A,$$ constants sum: $$-1728 - 3456 = -5184,$$ so $$576A - 5184 = 0 \Rightarrow 576A = 5184 \Rightarrow A = 9.$$ Then $$B = 16 - 9 = 7,$$ $$C = 192 - 24(9) = 192 - 216 = -24.$$ 8. **Rewrite the integral:** $$\int \frac{9}{x-18} + \frac{7}{x+6} - \frac{24}{(x+6)^2} \, dx = \int \frac{9}{x-18} \, dx + \int \frac{7}{x+6} \, dx - \int \frac{24}{(x+6)^2} \, dx.$$ 9. **Integrate term-by-term:** $$9 \ln|x-18| + 7 \ln|x+6| + 24 \frac{1}{x+6} + C,$$ note that integral of $-\frac{24}{(x+6)^2}$ is $24/(x+6)$ because: $$\int \frac{1}{(x+a)^2} dx = -\frac{1}{x+a} + C.$$ 10. **Final answer:** $$\int \frac{16x^2}{(x-18)(x+6)^2} \, dx = 9 \ln|x-18| + 7 \ln|x+6| + \frac{24}{x+6} + C.$$