Evaluation of a line integral using Green's Theorem where P, Q, and partial derivatives of P & Q are not continuous 

How can the author evaluate the below?
$$\oint_{C'}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$
Doesn't this contradict Theorem 9.12.1? P(0, 0) is undefined on region $R_2$ (corresponding to $C'$).
 A: The author is correct that he is able to do this, but he does not justify himself very well.  It is actually true that if two paths are homotopic (see wikipedia) through a region where the vector field is curl free, then the integrals over both paths will be equal.
I will draw some pictures to show you this is true only using version of greens theorem which you know. (sorry the picture is a bit dinky)

Green's theorem does apply to each of the regions $R_1,R_2,R_3,R_4$.  Since $Pdx+Qdy$ is curl free in each region, we know that the integral
$$\int_{bR_i} Pdx + Qdy = 0$$ holds for the oriented boundary of each $R_i$.  So
$$\int_{bR_1} Pdx + Qdy + \int_{bR_2} Pdx + Qdy + \int_{bR_3} Pdx + Qdy + \int_{bR_4} Pdx + Qdy = 0$$.
Now notice that there is some cancelation that occurs on the region of integrations!  For example, $R_1$ and $R_2$ both share the upper vertical line segment, but with different orientations.  In fact, looking at the second picture, you can see that all 4 line segments are present twice in the above sum of integrals, each time with opposite orientations.  So those integrals cancel out.  The remaining pieces are just those belonging to $C$ and $C'$, but with the opposite orientation!
$$\int_C Pdx +Qdy - \int_{C'} Pdx +Qdy = 0$$
$$\int_C Pdx +Qdy = \int_{C'} Pdx +Qdy$$
which is what your textbook claimed.
This is a powerful idea.  If you have some region of integration, you can often cut it up like this, and because the boundaries are oriented, you have cancellation on the interior.  In some sense this is the real reason for Green's theorem to begin with (It is true for rectangles by using the fundamental theorem twice, then decompose a region into rectangles and do this kind of thing to see the general result)
