Calculate $\int_C F·dr$ with $F=(e^y+ye^x, xe^y+e^x)$ where C is the path from $(a,0)$ to $(-a,0)$ over the upper half of the ellipse $b^2x^2+a^2y^2=a^2b^2$.
So $x=a\sin(t)$ and $y=bcos(t)$, then $r(t)=(a\sin(t), b\cos(t))$ and $r'(t)=(a\cos(t),-b\sin(t))$
Then $$F(t)=(e^{b\cos(t)}+b\cos(t)e^{a\sin(t)},a\sin(t)e^{b\cos(t)}+e^{a\sin(t)})$$
Then we have:
$$\int_0^\pi a\cos(t)e^{b\cos(t)}+abcos^2(t)e^{a\sin(t)}-absin^2(t)e^{b\cos(t)}-b\sin(t)e^{a\sin(t)} dt.$$
But the problem here is how to solve the integral, but we know that the answer is -2a.
You're making your life too hard -- this is a conservative vector field! Notice that $$ \vec{F}=\nabla G,\qquad G(x,y):=xe^y+ye^x. $$ So, EVERY route from $(a,0)$ to $(-a,0)$ will have the same integral, and that integral is precisely $$ G(-a, 0)-G(a,0)=(-ae^0+0e^{-a})-(ae^0+0e^a)=-2a. $$