1. **State the problem:** Calculate the line integral $\int_C (xy^2 + 1)\,dx + x(y^2 - 1)\,dy$ where $C$ is the line segment from $(2,4)$ to $(4,6)$. 2. **Parameterize the curve:** The line segment can be described by $y = x + 2$ since the slope $m = \frac{6-4}{4-2} = 1$ and passes through $(2,4)$. 3. **Express $dy/dx$:** Since $y = x + 2$, we have $\frac{dy}{dx} = 1$. 4. **Rewrite the integral:** The integral becomes $$\int_2^4 \left[(x y^2 + 1) + x(y^2 - 1) \frac{dy}{dx}\right] dx = \int_2^4 \left[(x (x+2)^2 + 1) + x ((x+2)^2 - 1) \cdot 1\right] dx$$ 5. **Simplify the integrand:** $(x (x+2)^2 + 1) + x ((x+2)^2 - 1) = x(x+2)^2 + 1 + x(x+2)^2 - x = 2x(x+2)^2 + 1 - x$ Expand $(x+2)^2 = x^2 + 4x + 4$, so $$2x(x^2 + 4x + 4) + 1 - x = 2x^3 + 8x^2 + 8x + 1 - x = 2x^3 + 8x^2 + 7x + 1$$ 6. **Integrate term-by-term:** $$\int_2^4 (2x^3 + 8x^2 + 7x + 1) dx = \left[\frac{2x^4}{4} + \frac{8x^3}{3} + \frac{7x^2}{2} + x\right]_2^4 = \left[\frac{x^4}{2} + \frac{8x^3}{3} + \frac{7x^2}{2} + x\right]_2^4$$ 7. **Evaluate at the bounds:** At $x=4$: $$\frac{4^4}{2} + \frac{8 \cdot 4^3}{3} + \frac{7 \cdot 4^2}{2} + 4 = \frac{256}{2} + \frac{8 \cdot 64}{3} + \frac{7 \cdot 16}{2} + 4 = 128 + \frac{512}{3} + 56 + 4$$ At $x=2$: $$\frac{2^4}{2} + \frac{8 \cdot 2^3}{3} + \frac{7 \cdot 2^2}{2} + 2 = \frac{16}{2} + \frac{8 \cdot 8}{3} + \frac{7 \cdot 4}{2} + 2 = 8 + \frac{64}{3} + 14 + 2$$ 8. **Calculate the difference:** $$\left(128 + \frac{512}{3} + 56 + 4\right) - \left(8 + \frac{64}{3} + 14 + 2\right) = (128 - 8) + \left(\frac{512}{3} - \frac{64}{3}\right) + (56 - 14) + (4 - 2)$$ $$= 120 + \frac{448}{3} + 42 + 2 = 164 + \frac{448}{3} = \frac{492}{3} + \frac{448}{3} = \frac{940}{3} \approx 313.33$$ **Final answer:** $$\boxed{\frac{940}{3} \approx 313.33}$$