1. The problem is to find the indefinite integral $$\int \frac{3x - 4}{2x - 4} \, dx$$. 2. To solve this, we use the method of algebraic manipulation and substitution. First, simplify the integrand if possible. 3. Rewrite the integrand: $$\frac{3x - 4}{2x - 4}$$. 4. Perform polynomial division or express the numerator in terms of the denominator: Let’s write $$3x - 4 = A(2x - 4) + B$$ for constants $$A$$ and $$B$$. 5. Expanding: $$3x - 4 = A(2x - 4) + B = 2Ax - 4A + B$$. 6. Equate coefficients: For $$x$$: $$3 = 2A \implies A = \frac{3}{2}$$. For constants: $$-4 = -4A + B \implies B = -4 + 4A = -4 + 4 \times \frac{3}{2} = -4 + 6 = 2$$. 7. So, $$\frac{3x - 4}{2x - 4} = \frac{3}{2} + \frac{2}{2x - 4}$$. 8. Now the integral becomes: $$\int \left( \frac{3}{2} + \frac{2}{2x - 4} \right) dx = \int \frac{3}{2} dx + \int \frac{2}{2x - 4} dx$$. 9. Integrate each term: - $$\int \frac{3}{2} dx = \frac{3}{2} x + C_1$$. - For $$\int \frac{2}{2x - 4} dx$$, use substitution: Let $$u = 2x - 4 \implies du = 2 dx \implies dx = \frac{du}{2}$$. 10. Substitute: $$\int \frac{2}{u} \times \frac{du}{2} = \int \frac{1}{u} du = \ln|u| + C_2 = \ln|2x - 4| + C_2$$. 11. Combine results: $$\int \frac{3x - 4}{2x - 4} dx = \frac{3}{2} x + \ln|2x - 4| + C$$, where $$C = C_1 + C_2$$ is the constant of integration. Final answer: $$\boxed{\int \frac{3x - 4}{2x - 4} dx = \frac{3}{2} x + \ln|2x - 4| + C}$$