Solve $\displaystyle \int_{A} \frac{x^2y + y^3 - y}{x^2 + y^2} \, dx + \frac{x^3 + xy^2 + x}{x^2 + y^2} \, dy $ , $A$ is the unit circle.
My attempt:
$\displaystyle \int_{A} \frac{x^2y + y^3 - y}{x^2 + y^2} \, dx + \frac{x^3 + xy^2 + x}{x^2 + y^2} \, dy =\int_A ydx+xdy-(\frac{y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy)=\displaystyle\int_A ydx+xdy+\int_A(\frac{y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy)$
First,using Green's theorem $\displaystyle\int_A ydx+xdy =0 $.
Second, $\int_A(\frac{y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy)$
Denote $x=\cos\theta,y=\sin\theta \implies dx=-\sin\theta,dy=\cos\theta$.
Therefore , $\int_A(\frac{y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy)=\int_0^{2\pi}-sin^2(\theta)+cos^2(\theta)=0$
$\displaystyle \int_{A} \frac{x^2y + y^3 - y}{x^2 + y^2} \, dx + \frac{x^3 + xy^2 + x}{x^2 + y^2} \, dy=0$
My answer isn't correct , can't find out what I am doing , appreciate any help.
It's just a sign mistake. You should have that $$\int_{A} \frac{x^2y + y^3 - y}{x^2 + y^2} \, \mathrm{d}x+ \frac{x^3 + xy^2 + x}{x^2 + y^2} \, \mathrm{d}y=I_1+I_2\, ,$$ where $$I_1=\int_A \left(y~\mathrm{d}x+x~\mathrm{d}y\right)\, ,$$ $$I_2=\int_A \left(\color{red}{-}\frac{y}{x^2+y^2}~\mathrm{d}x+\frac{x}{x^2+y^2}~\mathrm{d}y\right)\, .$$ As you correctly noted, $I_1=0$ by Green's theorem. You can compute $I_2$ either by using the same method you have tried, i.e. using the parametrisation $x=\cos(\theta)$, $y=\sin(\theta)$, or by simply noting that $$-\frac{y}{x^2+y^2}~\mathrm{d}x+\frac{x}{x^2+y^2}~\mathrm{d}y=\mathrm{d}\left(\arctan\frac{y}{x}\right).$$