Calculate the line integral $$ \int_\gamma \frac{y\,dx+(1-x)\,dy}{(x-1)^2+y^2} $$ where $\gamma$ is the ellipse $x^2 + 4y^2 = 4$ traversed two laps in positive direction.
So I have been given a solution but I don't understand it. It says that $F$ is not defined at $(1,0)$ but everywhere else it holds that $Q_x=P_y$, thus by Green's formula we can change the integration path to a small circle around $(1,0)$. Let $x = 1 + \epsilon\cos\theta$ and $y = \epsilon\sin\theta$. Inserted in the integral we get $-4\pi$.
So I don't understand how a small circle gives the desired result, and also if $x = 1 + \epsilon\cos\theta$ and $y = \epsilon\sin\theta$ then $$ \int^{4\pi}_0 \! (-\sin^2t + \cos^2t)\, dt = 0. $$
$F(x,y)=(P(x,y),Q(x,y))=(\frac{y}{(x-1)^2+y^2},\frac{1-x}{(x-1)^2+y^2})$