1. **State the problem:** We want to evaluate the integral $$\int \frac{9x^3 - 7x + 5}{x^3 - x^2} \, dx.$$\n\n2. **Perform long division:** Since the degree of numerator (3) is equal to degree of denominator (3), divide $$9x^3 - 7x + 5$$ by $$x^3 - x^2$$.\n\nDivide leading terms: $$\frac{9x^3}{x^3} = 9.$$\nMultiply divisor by 9: $$9(x^3 - x^2) = 9x^3 - 9x^2.$$\nSubtract: $$(9x^3 - 7x + 5) - (9x^3 - 9x^2) = 0x^3 + 9x^2 - 7x + 5.$$\n\nSo, $$\frac{9x^3 - 7x + 5}{x^3 - x^2} = 9 + \frac{9x^2 - 7x + 5}{x^3 - x^2}.$$\n\n3. **Factor the denominator of the remainder:** $$x^3 - x^2 = x^2(x - 1).$$\n\n4. **Set up partial fractions for the proper fraction:**\n$$\frac{9x^2 - 7x + 5}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}.$$\n\nMultiply both sides by $$x^2(x - 1)$$ to clear denominators:\n$$9x^2 - 7x + 5 = A x (x - 1) + B (x - 1) + C x^2.$$\n\n5. **Expand right side:**\n$$A x^2 - A x + B x - B + C x^2 = (A + C) x^2 + (B - A) x - B.$$\n\n6. **Equate coefficients:**\nFor $$x^2$$: $$9 = A + C,$$\nFor $$x$$: $$-7 = B - A,$$\nConstant: $$5 = -B.$$\n\n7. **Solve system:**\nFrom constant: $$B = -5.$$\nFrom $$x$$: $$-7 = -5 - A \Rightarrow A = -2.$$\nFrom $$x^2$$: $$9 = -2 + C \Rightarrow C = 11.$$\n\n8. **Rewrite partial fractions:**\n$$\frac{9x^2 - 7x + 5}{x^2(x - 1)} = \frac{-2}{x} + \frac{-5}{x^2} + \frac{11}{x - 1}.$$\n\n9. **Rewrite integral:**\n$$\int \frac{9x^3 - 7x + 5}{x^3 - x^2} \, dx = \int 9 \, dx + \int \left( \frac{-2}{x} + \frac{-5}{x^2} + \frac{11}{x - 1} \right) dx.$$\n\n10. **Integrate term-by-term:**\n$$\int 9 \, dx = 9x + C_1,$$\n$$\int \frac{-2}{x} \, dx = -2 \ln|x| + C_2,$$\n$$\int \frac{-5}{x^2} \, dx = \int -5 x^{-2} \, dx = -5 \left( \frac{x^{-1}}{-1} \right) + C_3 = \frac{5}{x} + C_3,$$\n$$\int \frac{11}{x - 1} \, dx = 11 \ln|x - 1| + C_4.$$\n\n11. **Combine all results:**\n$$\int \frac{9x^3 - 7x + 5}{x^3 - x^2} \, dx = 9x - 2 \ln|x| + \frac{5}{x} + 11 \ln|x - 1| + C,$$\nwhere $$C$$ is the constant of integration.