1. **State the problem:** We want to find the integral $$\int \frac{x + 2}{(x - 8)(x - 3)^2} \, dx.$$\n\n2. **Use partial fraction decomposition:** Since the denominator has linear factors $(x - 8)$ and $(x - 3)^2$, we express the integrand as\n$$\frac{x + 2}{(x - 8)(x - 3)^2} = \frac{A}{x - 8} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}.$$\nHere, $A$, $B$, and $C$ are constants to be determined.\n\n3. **Multiply both sides by the denominator to clear fractions:**\n$$x + 2 = A(x - 3)^2 + B(x - 8)(x - 3) + C(x - 8).$$\n\n4. **Expand the right side:**\n- Expand $(x - 3)^2 = x^2 - 6x + 9$.\n- Expand $(x - 8)(x - 3) = x^2 - 11x + 24$.\nSo,\n$$x + 2 = A(x^2 - 6x + 9) + B(x^2 - 11x + 24) + C(x - 8).$$\n\n5. **Group like terms:**\n$$x + 2 = (A + B)x^2 + (-6A - 11B + C)x + (9A + 24B - 8C).$$\n\n6. **Match coefficients with the left side:**\nSince the left side is $x + 2$, it can be written as $0x^2 + 1x + 2$. Equate coefficients:\n- For $x^2$: $A + B = 0$\n- For $x$: $-6A - 11B + C = 1$\n- For constant: $9A + 24B - 8C = 2$\n\n7. **Solve the system:**\nFrom $A + B = 0$, we get $B = -A$. Substitute into the other equations:\n- $-6A - 11(-A) + C = 1 \Rightarrow -6A + 11A + C = 1 \Rightarrow 5A + C = 1$\n- $9A + 24(-A) - 8C = 2 \Rightarrow 9A - 24A - 8C = 2 \Rightarrow -15A - 8C = 2$\n\nFrom $5A + C = 1$, we get $C = 1 - 5A$. Substitute into $-15A - 8C = 2$:\n$$-15A - 8(1 - 5A) = 2 \Rightarrow -15A - 8 + 40A = 2 \Rightarrow 25A - 8 = 2 \Rightarrow 25A = 10 \Rightarrow A = \frac{10}{25} = \frac{2}{5}.$$\n\nThen, $B = -A = -\frac{2}{5}$ and $C = 1 - 5A = 1 - 5 \times \frac{2}{5} = 1 - 2 = -1$.\n\n8. **Rewrite the integrand:**\n$$\frac{x + 2}{(x - 8)(x - 3)^2} = \frac{2/5}{x - 8} - \frac{2/5}{x - 3} - \frac{1}{(x - 3)^2}.$$\n\n9. **Integrate term by term:**\n$$\int \frac{2/5}{x - 8} \, dx = \frac{2}{5} \ln|x - 8| + C_1,$$\n$$\int -\frac{2/5}{x - 3} \, dx = -\frac{2}{5} \ln|x - 3| + C_2,$$\n$$\int -\frac{1}{(x - 3)^2} \, dx = \int -(x - 3)^{-2} \, dx = (x - 3)^{-1} + C_3 = \frac{1}{x - 3} + C_3.$$\n\n10. **Combine all results:**\n$$\int \frac{x + 2}{(x - 8)(x - 3)^2} \, dx = \frac{2}{5} \ln|x - 8| - \frac{2}{5} \ln|x - 3| + \frac{1}{x - 3} + C,$$\nwhere $C$ is the constant of integration.\n