$\oint_C (x^3+x)dx+(1+y^2+y^3)dy$ , where $C=\{(x(t),y(t))| x(t)=2+3\cos t ,y(t)=5+4\sin t , 0\leq t<2\pi\}$

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Find the value of the integral $\oint_C (x^3+x)dx+(1+y^2+y^3)dy$ , where $C=\{(x(t),y(t))| x(t)=2+3\cos t ,y(t)=5+4\sin t , 0\leq t<2\pi\}$

My work :

I think the answer is $0$

$C: (\frac{x-2}{3})^2+(\frac{y-5}{4})^2=1$ , an eclipse . Note that $(x^3+x)dx+(1+y^2+y^3)dy=d(\frac{x^4}{4}+\frac{x^2}{2}+y+\frac{y^3}{3}+\frac{y^4}{4})$ .

So the line integral is integrated over a curve where initial point and end point coincide and the integrand is an exact differential. So the line line integral$=0$