Standard way:
Notice the interior of the area enclosed the curve $C$ has a singularity of the vector field, i.e., $\displaystyle \frac{\partial Q}{\partial x}$ or $\displaystyle \frac{\partial P}{\partial y}$ is not continuous.
Standard procedure is to cut a hole with a small radius $r$ centered at this singularity , in this example it is $(0,0)$, such that outside this small disk of radius $r$ there is no singularity (See the picture above).
Denote $C'$ the boundary of this smaller disk $\Omega'$, rotating clockwisely. Then by $P = x/(x^2+y^2)$, $Q = y/(x^2+y^2)$:
\begin{align}
&\oint_{C} \frac{xdy-ydx}{x^2+y^2} + \oint_{C'} \frac{xdy-ydx}{x^2+y^2}
\\
=&
\int_{\Omega\backslash \Omega'}\left\{ \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right)+\frac{\partial}{\partial y}\left(\frac{y}{x^2+y^2}\right)\right\} dx dy = 0.
\end{align}
Now parametrize the curve $C'$ using $t$, by letting $x = r\cos t$, $y = r\sin t$, so $dx = -r\sin t\, dt$, $dy = r \cos t \,dt$, $t$ from $2\pi$ to $0$ (clockwise). The integral becomes:
\begin{align}
& \oint_{C} \frac{xdy-ydx}{x^2+y^2} = \oint_{C'}\frac{y }{x^2+y^2}dx-\frac{x}{x^2+y^2}dy
\\
=& \int_{2\pi}^0 \left(\frac{r\sin t\, }{r^2\cos^2 t + r^2\sin^2 t} (-r\sin t)
- \frac{r\cos t\,}{r^2\cos^2 t + r^2\sin^2 t}r \cos t \right)dt
=2\pi.
\end{align}
Cheating way: since Muphrid mentioned the delta function, we can evaluate using divergence theorem (a little cheated): Notice $F = \nabla \phi$, where $\phi = \big(\ln(x^2+y^2)\big)$.
$$
\mathrm{Flux} = \int_{\partial \Omega} F\cdot n \,ds = \frac{1}{2}\int_{\partial \Omega} \nabla \phi \cdot n \,ds = \frac{1}{2} \int_{\Omega} \Delta \phi\,dx = \frac{1}{2} \int_{\Omega} 4\pi \delta_0(x) = 2\pi.
$$
Cheating way #2 : The domain has winding number being 1, so by definition:
$$
\frac{1}{2\pi} \oint_C \,\frac{x}{x^2+y^2}\,dy - \frac{y}{x^2+y^2}\,dx= 1.
$$
