1. **State the problem:** We need to evaluate the integral $$\int \frac{z^2 + 20z - 15}{z^3 + 4z^2 - 5z} \, dz.$$\n\n2. **Factor the denominator:** The denominator is a cubic polynomial. Factor it to simplify the integral.\n$$z^3 + 4z^2 - 5z = z(z^2 + 4z - 5)$$\nFactor the quadratic: $$z^2 + 4z - 5 = (z + 5)(z - 1)$$\nSo denominator factors as $$z(z + 5)(z - 1).$$\n\n3. **Set up partial fractions:** We express the integrand as\n$$\frac{z^2 + 20z - 15}{z(z + 5)(z - 1)} = \frac{A}{z} + \frac{B}{z + 5} + \frac{C}{z - 1}.$$\n\n4. **Multiply both sides by the denominator:**\n$$z^2 + 20z - 15 = A(z + 5)(z - 1) + B z (z - 1) + C z (z + 5).$$\n\n5. **Expand each term:**\n$$A(z^2 + 4z - 5) + B(z^2 - z) + C(z^2 + 5z).$$\n\n6. **Combine like terms:**\n$$A z^2 + 4 A z - 5 A + B z^2 - B z + C z^2 + 5 C z = (A + B + C) z^2 + (4 A - B + 5 C) z - 5 A.$$\n\n7. **Match coefficients with the numerator:**\nFor $$z^2$$: $$1 = A + B + C$$\nFor $$z$$: $$20 = 4 A - B + 5 C$$\nFor constant: $$-15 = -5 A$$\n\n8. **Solve for A:**\n$$-15 = -5 A \implies A = 3.$$\n\n9. **Substitute A = 3 into the other equations:**\n$$1 = 3 + B + C \implies B + C = -2,$$\n$$20 = 4(3) - B + 5 C = 12 - B + 5 C.$$\nRearranged: $$- B + 5 C = 8.$$\n\n10. **Solve the system:**\nFrom $$B + C = -2$$, we get $$B = -2 - C.$$\nSubstitute into $$- B + 5 C = 8$$:\n$$-(-2 - C) + 5 C = 8 \implies 2 + C + 5 C = 8 \implies 6 C = 6 \implies C = 1.$$\nThen $$B = -2 - 1 = -3.$$\n\n11. **Rewrite the integral:**\n$$\int \left( \frac{3}{z} + \frac{-3}{z + 5} + \frac{1}{z - 1} \right) dz = \int \frac{3}{z} dz - \int \frac{3}{z + 5} dz + \int \frac{1}{z - 1} dz.$$\n\n12. **Integrate each term:**\n$$3 \ln|z| - 3 \ln|z + 5| + \ln|z - 1| + C,$$\nwhere $$C$$ is the constant of integration.\n\n**Final answer:**\n$$\boxed{3 \ln|z| - 3 \ln|z + 5| + \ln|z - 1| + C}.$$