Answer
$\int_Cydx+(x+y^{2})dy=0$
Work Step by Step
Rewrite the equation of the ellipse as
$(\frac{x}{3})^{2}+(\frac{y}{2})^{2}=1$
We can parameterize the ellipse as
$C:x=3cost,y=2sint$
Starting from $(3,0)$ when we complete one round of the ellipse, $t$ increases from $0$ to $2\pi$
Note that $dx=-3sintdt$ and $dy=2costdt$
Therefore,
$\int_Cydx+(x+y^{2})dy=\int_{0}^{2\pi}2sint(-3sint)dt+(3cost+4sin^{2}t)(2cost)dt$
$=\int_{0}^{2\pi}6cos2t+8(1-cos^{2}t)(cost)dt$
$=\int_{0}^{2\pi}6cos2t+8cost-8cos^{3}tdt$
Use formula: $4cos^{3}\theta=3cos\theta-cos3\theta$
$=\int_{0}^{2\pi}6cos2t+8cost-6cost+2cos3tdt$
$=0$
Hence, $\int_Cydx+(x+y^{2})dy=0$
