1. **State the problem**: We need to evaluate the definite integral $$\int_1^3 \frac{4x^3 - 2x + 1}{x^2} \, dx.$$\n\n2. **Simplify the integrand** by dividing each term by $$x^2$$:\n$$\frac{4x^3}{x^2} - \frac{2x}{x^2} + \frac{1}{x^2} = 4x - 2x^{-1} + x^{-2}.$$\n\n3. **Rewrite the integral** with the simplified integrand:\n$$\int_1^3 (4x - 2x^{-1} + x^{-2}) \, dx.$$\n\n4. **Integrate each term** separately using power rules:\n- Integral of $$4x$$ is $$4 \cdot \frac{x^2}{2} = 2x^2$$\n- Integral of $$-2x^{-1}$$ is $$-2 \ln|x|$$\n- Integral of $$x^{-2}$$ is $$\frac{x^{-1}}{-1} = -x^{-1}$$\n\nSo, the antiderivative is:\n$$F(x) = 2x^2 - 2 \ln|x| - x^{-1} + C.$$\n\n5. **Evaluate the definite integral** from 1 to 3:\n$$\int_1^3 \frac{4x^3 - 2x + 1}{x^2} \, dx = F(3) - F(1) = \left(2(3)^2 - 2 \ln 3 - \frac{1}{3}\right) - \left(2(1)^2 - 2 \ln 1 - 1\right).$$\n\n6. **Calculate values**:\n- $$F(3) = 2 \times 9 - 2 \ln 3 - \frac{1}{3} = 18 - 2 \ln 3 - \frac{1}{3}.$$\n- $$F(1) = 2 - 0 - 1 = 1$$ since $$\ln 1 = 0.$$\n\n7. **Subtract**:\n$$18 - 2 \ln 3 - \frac{1}{3} - 1 = 17 - 2 \ln 3 - \frac{1}{3} = \frac{51}{3} - \frac{1}{3} - 2 \ln 3 = \frac{50}{3} - 2 \ln 3.$$\n\n**Final answer**:\n$$\int_1^3 \frac{4x^3 - 2x + 1}{x^2} \, dx = \frac{50}{3} - 2 \ln 3.$$