## Calculus 8th Edition

$1.72599$
Here, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ This implies that $ds=\sqrt{(3)^2+(2t)^2+(4t^3)^2}dt=\sqrt {9+4t^2+16t^{16}} dt$ Further, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_{-1}^1 (t^4) \cdot \ln [(1+3t)+(2+t^2)] \cdot (\sqrt {9+4t^2+16t^{16}} dt)$ Hence, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=1.72599$ (Using calculator)