1. **State the problem:** We want to evaluate the integral $$\int \frac{7x^3 - 8x + 5}{x^3 - x^2} \, dx$$ by first performing long division, then expressing the proper fraction as partial fractions, and finally integrating. 2. **Perform long division:** Divide the numerator by the denominator. Divide $$7x^3$$ by $$x^3$$ to get $$7$$. Multiply $$7$$ by the denominator $$x^3 - x^2$$ to get $$7x^3 - 7x^2$$. Subtract this from the numerator: $$\begin{aligned} (7x^3 - 8x + 5) - (7x^3 - 7x^2) &= 7x^3 - 8x + 5 - 7x^3 + 7x^2 \\ &= 7x^2 - 8x + 5 \end{aligned}$$ So the division gives: $$\frac{7x^3 - 8x + 5}{x^3 - x^2} = 7 + \frac{7x^2 - 8x + 5}{x^3 - x^2}$$ 3. **Factor the denominator of the proper fraction:** $$x^3 - x^2 = x^2(x - 1)$$ 4. **Set up partial fractions:** $$\frac{7x^2 - 8x + 5}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}$$ Multiply both sides by $$x^2(x - 1)$$: $$7x^2 - 8x + 5 = A x (x - 1) + B (x - 1) + C x^2$$ 5. **Expand the right side:** $$A x^2 - A x + B x - B + C x^2 = (A + C) x^2 + (B - A) x - B$$ 6. **Equate coefficients:** For $$x^2$$: $$7 = A + C$$ For $$x$$: $$-8 = B - A$$ For constant term: $$5 = -B$$ 7. **Solve the system:** From constant term: $$B = -5$$ From $$x$$ term: $$-8 = -5 - A \Rightarrow A = -5 + 8 = 3$$ From $$x^2$$ term: $$7 = 3 + C \Rightarrow C = 4$$ 8. **Rewrite the integrand:** $$\frac{7x^3 - 8x + 5}{x^3 - x^2} = 7 + \frac{3}{x} - \frac{5}{x^2} + \frac{4}{x - 1}$$ 9. **Integrate term-by-term:** $$\int \left(7 + \frac{3}{x} - \frac{5}{x^2} + \frac{4}{x - 1}\right) dx = \int 7 \, dx + \int \frac{3}{x} \, dx - \int \frac{5}{x^2} \, dx + \int \frac{4}{x - 1} \, dx$$ Calculate each integral: $$7x + 3 \ln|x| + 5 \frac{1}{x} + 4 \ln|x - 1| + C$$ (Note: Integral of $$-5/x^2 = -5 \int x^{-2} dx = -5 (-x^{-1}) = 5/x$$) **Final answer:** $$\int \frac{7x^3 - 8x + 5}{x^3 - x^2} \, dx = 7x + 3 \ln|x| + \frac{5}{x} + 4 \ln|x - 1| + C$$