Problem
Compute $$ \oint_L\frac{xdy-ydx}{4x^2+y^2} $$ where $L$ is a circle centered at $(1, 0)$ with a radius of $R > 1$, and the direction of $L$ is counterclock-wise.
Solution
To by pass the singular point at $(0, 0)$ which is inside the circle $L$ surrounds, adding an auxiliary curve: $$ C:4x^2+y^2 = \delta^2 $$ The integral can then be computed by applying Green's theorem. However, different choices of the direction of $C$ produce different answers. If we choose counterclock-wise: $$ \oint_L\frac{xdy-ydx}{4x^2+y^2}=\oint_{L+C}-\oint_{C}=-\oint_{C}=-\pi $$ ， if we choose clockwise: $$ \oint_L\frac{xdy-ydx}{4x^2+y^2}=\oint_{L+C^{-1}}-\oint_{C^{-1}}=\oint_{C}=\pi $$
What I got wrong here?
The "pitfall" here is that Green's theorem only works with positively oriented boundaries/curves, which means that you need the correct orientation for $C$, which should be the opposite of that of $L$ (see for example this answer for the reasoning behind that: https://math.stackexchange.com/a/141869/1104384), because you want and need the union of $L$ and $C$ to be positively oriented.
In this case, that means that $C$ must be oriented clockwise, hence you get $+ \pi$.