1. **State the problem:** We need to solve the integral $$\int \frac{x^3+8}{(x^2-1)(x-2)} \, dx.$$\n\n2. **Rewrite the denominator:** Note that $$x^2-1 = (x-1)(x+1),$$ so the integral becomes $$\int \frac{x^3+8}{(x-1)(x+1)(x-2)} \, dx.$$\n\n3. **Use partial fraction decomposition:** We express $$\frac{x^3+8}{(x-1)(x+1)(x-2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-2}.$$\n\n4. **Multiply both sides by the denominator:** $$x^3+8 = A(x+1)(x-2) + B(x-1)(x-2) + C(x-1)(x+1).$$\n\n5. **Expand each term:**\n- $$A(x+1)(x-2) = A(x^2 - x - 2) = A x^2 - A x - 2A,$$\n- $$B(x-1)(x-2) = B(x^2 - 3x + 2) = B x^2 - 3B x + 2B,$$\n- $$C(x-1)(x+1) = C(x^2 - 1) = C x^2 - C.$$\n\n6. **Combine all terms:** $$x^3 + 8 = (A + B + C) x^2 + (-A - 3B) x + (-2A + 2B - C).$$\n\n7. **Match coefficients with the left side:**\n- Coefficient of $$x^3$$: Left side is 1, right side has none, so we add $$D x^3$$ term or realize we need to adjust. Since denominator is degree 3 and numerator degree 3, perform polynomial division first.\n\n8. **Polynomial division:** Divide $$x^3 + 8$$ by $$x^3 - 2x^2 - x + 2$$ (since $$(x-1)(x+1)(x-2) = x^3 - 2x^2 - x + 2$$).\n\n- Divide leading terms: $$x^3 / x^3 = 1.$$\n- Multiply divisor by 1: $$x^3 - 2x^2 - x + 2.$$\n- Subtract: $$(x^3 + 8) - (x^3 - 2x^2 - x + 2) = 0 + 2x^2 + x + 6.$$\n\n9. **Rewrite integral:** $$\int \frac{x^3 + 8}{(x-1)(x+1)(x-2)} dx = \int 1 \, dx + \int \frac{2x^2 + x + 6}{(x-1)(x+1)(x-2)} dx.$$\n\n10. **Now decompose:** $$\frac{2x^2 + x + 6}{(x-1)(x+1)(x-2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-2}.$$\n\n11. **Multiply both sides by denominator:** $$2x^2 + x + 6 = A(x+1)(x-2) + B(x-1)(x-2) + C(x-1)(x+1).$$\n\n12. **Expand:**\n- $$A(x^2 - x - 2) = A x^2 - A x - 2A,$$\n- $$B(x^2 - 3x + 2) = B x^2 - 3B x + 2B,$$\n- $$C(x^2 - 1) = C x^2 - C.$$\n\n13. **Combine:** $$2x^2 + x + 6 = (A + B + C) x^2 + (-A - 3B) x + (-2A + 2B - C).$$\n\n14. **Match coefficients:**\n- For $$x^2$$: $$2 = A + B + C,$$\n- For $$x$$: $$1 = -A - 3B,$$\n- For constant: $$6 = -2A + 2B - C.$$\n\n15. **Solve the system:**\nFrom $$1 = -A - 3B$$, we get $$A = -1 - 3B.$$\n\nSubstitute into $$2 = A + B + C$$:\n$$2 = (-1 - 3B) + B + C = -1 - 2B + C \Rightarrow C = 3 + 2B.$$\n\nSubstitute $$A$$ and $$C$$ into $$6 = -2A + 2B - C$$:\n$$6 = -2(-1 - 3B) + 2B - (3 + 2B) = 2 + 6B + 2B - 3 - 2B = (2 - 3) + (6B + 2B - 2B) = -1 + 6B.$$\n\nSo, $$6 = -1 + 6B \Rightarrow 6B = 7 \Rightarrow B = \frac{7}{6}.$$\n\nThen $$A = -1 - 3 \times \frac{7}{6} = -1 - \frac{7}{2} = -\frac{9}{2}.$$\n\nAnd $$C = 3 + 2 \times \frac{7}{6} = 3 + \frac{7}{3} = \frac{16}{3}.$$\n\n16. **Rewrite integral:**\n$$\int 1 \, dx + \int \left( \frac{-\frac{9}{2}}{x-1} + \frac{7/6}{x+1} + \frac{16/3}{x-2} \right) dx.$$\n\n17. **Integrate term by term:**\n$$x - \frac{9}{2} \ln|x-1| + \frac{7}{6} \ln|x+1| + \frac{16}{3} \ln|x-2| + C,$$ where $$C$$ is the constant of integration.\n\n**Final answer:**\n$$\boxed{x - \frac{9}{2} \ln|x-1| + \frac{7}{6} \ln|x+1| + \frac{16}{3} \ln|x-2| + C}.$$