Interesting line integral using green's theorem Find $$\int_C \frac{2x^3+2xy^2-2y}{ x^2+y^2} \, dx+\frac{2y^3+2x^2y+2x}{ x^2+y^2} \, dy$$
Where $C$ is any simple closed loop which contains the origin.

What I figured out
I cannot use the direct version of Green's theorem.
I know that there is another version of greens theorem which is as follows :
$$\oint_{C+S} P \,dx + Q\,dy  = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA,$$
where $C$ and $S$ are oriented opposite to each other and $D$ is the region enclosed between the two loops.
Using that I found
\begin{align*}
\int_{C+S} \left( 2x-\frac{2y}{ x^2+y^2}\right) \, dx+\left(2y+\frac{2x}{ x^2+y^2} \right) \, dy
&= \iint_D \left[\left(2+\frac{4xy}{(x^2+y^2)^2} \right)-\left(2-\frac{4xy}{(x^2+y^2)^2} \right) \right]dA \\
&=\iint_D \frac{8xy}{(x^2+y^2)^2} dA
\end{align*}

The region inside any closed loop can be defined in polar coordinates as $$\{(r, \theta)\in D \mid 0\leq r\leq r(\theta), \, 0\leq \theta \leq 2\pi \}$$
In polar coordinates the integral becomes $$ =\iint_D \frac{8xy}{(x^2+y^2)^2} dA = \int_0^{2\pi}\!\!\int_0^{r(\theta)} \frac{4\sin2\theta}{r} dr\,d\theta $$
Oops, I am again stuck; it's not zero.

You can write your integral as
$$2\int_{C} \left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) + \int_{C} \left( 2x\ dx + 2y\ dy\right)\ ,$$
the second form is exact, so its contribution is 0, while for the first for you can  do the following. Note that $$\left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) =d\ \arctan\left(\frac{y}{x}\right)\ ,$$
which in polar coordinates can be written as $d\theta$. Your integral is thus just
$$-2\int_{C} d\theta = \mp 4\pi$$
where the sign depend on the orientation of the curve you choose: $-$ if it's counterclockwise, $+$ if clockwise. 
