1. **State the problem:** We need to find the integral $$\int \frac{2x^3 + 5x^2 + 3x - 5}{2x - 1} \, dx$$. 2. **Use polynomial division:** Since the degree of the numerator (3) is higher than the denominator (1), divide $$2x^3 + 5x^2 + 3x - 5$$ by $$2x - 1$$. 3. **Perform the division:** - Divide the leading term: $$\frac{2x^3}{2x} = x^2$$. - Multiply divisor by $$x^2$$: $$x^2(2x - 1) = 2x^3 - x^2$$. - Subtract: $$ (2x^3 + 5x^2) - (2x^3 - x^2) = 6x^2$$. - Bring down $$3x$$. - Divide $$6x^2$$ by $$2x$$: $$3x$$. - Multiply divisor by $$3x$$: $$3x(2x - 1) = 6x^2 - 3x$$. - Subtract: $$ (6x^2 + 3x) - (6x^2 - 3x) = 6x$$. - Bring down $$-5$$. - Divide $$6x$$ by $$2x$$: $$3$$. - Multiply divisor by $$3$$: $$3(2x - 1) = 6x - 3$$. - Subtract: $$ (6x - 5) - (6x - 3) = -2$$. So the division gives: $$\frac{2x^3 + 5x^2 + 3x - 5}{2x - 1} = x^2 + 3x + 3 + \frac{-2}{2x - 1}$$. 4. **Rewrite the integral:** $$\int \frac{2x^3 + 5x^2 + 3x - 5}{2x - 1} \, dx = \int (x^2 + 3x + 3) \, dx + \int \frac{-2}{2x - 1} \, dx$$. 5. **Integrate each term:** - $$\int (x^2 + 3x + 3) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + 3x + C_1$$. - For $$\int \frac{-2}{2x - 1} \, dx$$, use substitution: Let $$u = 2x - 1$$, then $$du = 2 dx$$, so $$dx = \frac{du}{2}$$. 6. **Substitute and integrate:** $$\int \frac{-2}{u} \cdot \frac{du}{2} = \int \frac{-2}{u} \cdot \frac{1}{2} du = \int \frac{-1}{u} du = -\ln|u| + C_2 = -\ln|2x - 1| + C_2$$. 7. **Combine results:** $$\int \frac{2x^3 + 5x^2 + 3x - 5}{2x - 1} \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + 3x - \ln|2x - 1| + C$$. **Final answer:** $$\boxed{\frac{x^3}{3} + \frac{3x^2}{2} + 3x - \ln|2x - 1| + C}$$