<s>[INST] Line integral $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ Solve $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ =? 
K: $x^2+y^2=1$ is oriented positively.

My attempt:
$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy = |x=cost, y=sint,x'=- sint,y'=cost | $ = *
I put for $dx = -sint,dy=cost$
$= \int_0^{2\pi} \frac{sint}{cost^2+sint^2} -sint -\frac{cost}{cost^2+sint^2} cost  dt=  -\int_0^{2\pi} 1 dt = - 2 \pi $
We can't use Green's theorem because $P(x,y)$ and $Q(x,y)$ aren't continuous and their diveratives aren't continuous at point. $(0,0)$?
 [/INST] Seems ok. You can't use Green-Stokes because of the reason you correctly stated. Another way to see this result is to note that if $z = x+iy$ and ${\rm d}z = {\rm d}x+i\,{\rm d}y$, then $$\frac{y}{x^2+y^2}\,{\rm d}x - \frac{x}{x^2+y^2}\,{\rm d}y = {\rm Im}\left(-\frac{{\rm d}z}{z}\right),$$and so $$ -{\rm Im} \int_{\Bbb S^1}\frac{{\rm d}z}{z} = -{\rm Im}(2\pi i) = -2\pi.$$
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