Prove that ${x dy - y d x\over x^2 + y^2}$ is not exact Please can someone verify my proof that
$$\psi = {x dy - y d x\over x^2 + y^2}$$
is not exact?
Here is my work:
If $\psi$ was exact there would exist $f:\mathbb R^2 \setminus \{0\} \to \mathbb R$ such that $df = f_x dx + f_y dy = \psi$. Here $f_x = {-y \over x^2 + y^2}$ and $f_y = {x \over x^2 + y^2}$.
It would hold true that
$$\int f_x dx = \int f_y dy = f.$$
So I calculate these integrals:
$$ \int f_x dx = -{1\over 2}\log(x^2 + y^2)$$
and
$$ \int f_y dy = {1\over 2}\log(x^2 + y^2)$$
It is clear that these cannot be equal therefore $\psi$ is not exact.
Edit
The first thing I had tried (it did not work) was to calculate the integral along a closed curve:
$$ \int_{S^1}\psi = \int_{S^1} {x \over x^2 + y^2} dy - \int_{S^1} {y \over x^2 + y^2} dx = \int_{S^1} x dy - \int_{S^1} y dx= x \int_{S^1}  dy - y \int_{S^1}  dx = 0$$
since $\int_{S^1}dx = 0$.

By Green's theorem, 
$$\int_{S^1} \psi = \int_{S^1} x\, dy - y\, dx = \iint_{D^2} \text{div}(\langle x,y\rangle)\, dA = 2\cdot \text{Area}(D^2) = 2\pi$$
Alternatively, parametrize $S^1$ by setting $x = \cos(t)$, $y = \sin(t)$, $0 \le t \le 2\pi$. Then $$\int_{S^1} \psi = \int_0^{2\pi} (\cos(t)\cdot \cos(t) - \sin(t)\cdot (-\sin(t)))\, dt = \int_0^{2\pi} (\cos^2(t) + \sin^2(t))\, dt = 2\pi$$
Either way, $\int_{S^1} \psi \neq 0$.
