Evaluating the line integral $\oint\limits_C \ {x-y\over x^2+y^2}dx+{x+y\over x^2+y^2}dy$ over a closed path $C$ 
Let $C$ be the closed path formed by the line $x=2$ and the parabola $y^2=2(x+2)$. Find $$\oint\limits_C \ {x-y\over x^2+y^2}dx+{x+y\over x^2+y^2}dy.$$

First I tried to find this integral for the parabola, with the parametrization $t \mapsto ({t^2\over 2}-2,t)$ but the resulting integral is very difficult to solve, and I can't think of any other parametrization, How can I solve this problem?
 A: Hint Computing gives that, if we (as usual) write the integral as $\oint_C P \,dx + Q \,dy$, then
$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0,$$
so it is tempting to apply Green's Theorem immediately. But, it does not apply here, as the $P$ and $Q$ are not defined on the region $R$ bounded by $C$, and in particular at $(0, 0)$.
We can, however, use Green's Theorem in a somewhat sneakier way to simply our computation: First, note that part of what makes the line integral tricky is the $x^2 + y^2$ that appears in the denomination, which, for a typical parameterization, will become rather complicated. This quantity is constant, however, on any circle centered at the origin, so it would simplify the problem dramatically if we could replace the given curve with such a circle.
We can do this as follows: Let $B$ denote a closed ball centered at the origin with radius small enough that it $B$ is contained inside (the interior) of $R$. Now, the (oriented) boundary of $R - B$ is the (appropriately) oriented union of $C$ and the circle $\partial B$, and $P, Q$ are defined on $R - B$ so we can apply Green's Theorem: On one hand, we have
$$\oint_{\partial(R - B)} P \,dx + Q \,dy = \iint_{R - B} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = \iint_{R - B} (0) \,dA = 0,$$ and on the other, decomposing the line integral gives
$$\oint_{\partial(R - B)} P \,dx + Q \,dy = \oint_C P \,dx + Q \,dy - \oint_{\partial B} P \,dx + Q \,dy.$$
So, rearranging gives
$$\oint_C P \,dx + Q \,dy = \oint_{\partial B} P \,dx + Q \,dy.$$
But the integral on the l.h.s. was the original integral we wanted to compute, and the integral on the r.h.s. is easier to compute: For example, if we denote the radius of $B$ by $r$, using the standard parameterization $\gamma(t) = (r \cos t, r \sin t)$ of $\partial B$ gives
$$P \, dx = \frac{x - y}{x^2 + y^2} dx = \frac{r \cos t - r \sin t}{r^2} d(r \cos t) = -\sin t (\cos t - \sin t).$$
