1. **State the problem:** We need to evaluate the definite integral $$\int_1^2 \frac{4x^2 + 2x - 5}{(x + 1)(2x - 1)} \, dx.$$\n\n2. **Rewrite the integrand:** The denominator is already factored as $(x+1)(2x-1)$. We want to express the numerator as a combination of these factors to use partial fractions.\n\n3. **Set up partial fractions:** Assume $$\frac{4x^2 + 2x - 5}{(x + 1)(2x - 1)} = \frac{A}{x+1} + \frac{B}{2x-1}.$$\nMultiply both sides by $(x+1)(2x-1)$ to clear denominators:\n$$4x^2 + 2x - 5 = A(2x - 1) + B(x + 1).$$\n\n4. **Expand and collect terms:**\n$$4x^2 + 2x - 5 = 2Ax - A + Bx + B = (2A + B)x + (-A + B).$$\n\n5. **Match coefficients:** The left side has a $4x^2$ term, but the right side does not. This means our assumption of simple partial fractions is insufficient because the numerator degree is equal to denominator degree. We must perform polynomial division first.\n\n6. **Polynomial division:** Divide $4x^2 + 2x - 5$ by $(x+1)(2x-1) = 2x^2 + x - 1$.\n\n- Leading term division: $4x^2 / 2x^2 = 2$.\n- Multiply divisor by 2: $2(2x^2 + x - 1) = 4x^2 + 2x - 2$.\n- Subtract: $(4x^2 + 2x - 5) - (4x^2 + 2x - 2) = -3$.\n\nSo, $$\frac{4x^2 + 2x - 5}{(x+1)(2x-1)} = 2 + \frac{-3}{(x+1)(2x-1)}.$$\n\n7. **Rewrite integral:**\n$$\int_1^2 \frac{4x^2 + 2x - 5}{(x+1)(2x-1)} dx = \int_1^2 2 \, dx + \int_1^2 \frac{-3}{(x+1)(2x-1)} \, dx.$$\n\n8. **Partial fractions for remainder:**\n$$\frac{-3}{(x+1)(2x-1)} = \frac{A}{x+1} + \frac{B}{2x-1}.$$\nMultiply both sides by denominator:\n$$-3 = A(2x - 1) + B(x + 1) = (2A + B)x + (-A + B).$$\n\n9. **Match coefficients:**\n- Coefficient of $x$: $0 = 2A + B$\n- Constant term: $-3 = -A + B$\n\nFrom $0 = 2A + B$, we get $B = -2A$. Substitute into second equation:\n$$-3 = -A + (-2A) = -3A \implies A = 1.$$\nThen $B = -2(1) = -2$.\n\n10. **Rewrite remainder:**\n$$\frac{-3}{(x+1)(2x-1)} = \frac{1}{x+1} - \frac{2}{2x-1}.$$\n\n11. **Integral becomes:**\n$$\int_1^2 2 \, dx + \int_1^2 \frac{1}{x+1} \, dx - \int_1^2 \frac{2}{2x-1} \, dx.$$\n\n12. **Integrate each term:**\n- $$\int_1^2 2 \, dx = 2x \Big|_1^2 = 2(2) - 2(1) = 4 - 2 = 2.$$\n- $$\int_1^2 \frac{1}{x+1} \, dx = \ln|x+1| \Big|_1^2 = \ln 3 - \ln 2 = \ln \frac{3}{2}.$$\n- For $$\int_1^2 \frac{2}{2x-1} \, dx,$$ let $u = 2x - 1$, then $du = 2 dx$, so $dx = \frac{du}{2}$.\n\nRewrite integral:\n$$\int_1^2 \frac{2}{2x-1} \, dx = \int_{u=1}^{u=3} \frac{2}{u} \cdot \frac{du}{2} = \int_1^3 \frac{1}{u} \, du = \ln|u| \Big|_1^3 = \ln 3 - \ln 1 = \ln 3.$$\n\n13. **Combine results:**\n$$2 + \ln \frac{3}{2} - \ln 3 = 2 + \ln \left( \frac{3}{2} \cdot \frac{1}{3} \right) = 2 + \ln \frac{1}{2} = 2 - \ln 2.$$\n\n**Final answer:** $$\boxed{2 - \ln 2}.$$