If $\gamma$ is a closed path, then $\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\in2\pi\mathbb{Z}$ As the title says:
Let $\gamma$ be a smooth closed path that avoids $(0,0)$. I need to show that the line integral: $$\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ is a multiple of $2\pi$.
Note that $\gamma$ is allowed to self-intersect itself.

Intuitively this value should represent the "number of rotations around $(0,0)$", as they are counted with orientation, but this intuition is all I have.

Any ideas?
Take $\gamma : [0,1] \to \mathbb C$ a complex valued function, so that the quotient $\gamma'/\gamma$ is exactly what you obtain when replacing $\gamma$ into
$$\frac{dz}{z} = \frac{x-iy}{x^2+y^2}(dx+i dy)$$
and multiplying out. Thus, consider the function $h(t) = \int_0^t \frac{\gamma'(s)}{\gamma(s)}ds$. Set $H(t) =\gamma(t) \exp(-h(t))$, and observe that $H'(t) = 0$. This means $H(t)$ is constant, and then that $H(1) = H(0)$. All in all, we have $\gamma(1) \exp (-h(1)) = \gamma(0) \exp (-h(0))$
which gives that $\exp h(1) = 1$ or, what is the same, that $h(1) = 2\pi i n$ with $n\in \mathbb Z$.

Remark: this elementary argument is to be found in Apostol's Mathematical Analysis.
Write $x=rsin(u), y=rsin(u)$, the form  $\alpha={{-y}\over{x^2+y^2}}dx+{{x}\over{x^2+y^2}}dy=du$ then take a circle $C$ contained in $\gamma$ and apply Stokes formula, since $d\alpha=0$, you obtain $\int_C\alpha=\int_{\gamma}\alpha=2\pi$.