I got this line integral: $$ \int \frac{(x+y)dx-(x-y)dy}{x^2+y^2} $$ Curve is a circle $$ x=a\cos(t) $$ $$ y=a\sin(t) $$ $$ 0 \leq t \leq 2\pi $$
I first tried to solve it directly and got $ -2a^2\pi $ as a result, however after that I calculated partial derivatives of M and N: $$ \frac{\partial M}{\partial y}=\frac{x^2-y^2-2xy}{(x^+y^2)^2} $$ $$ \frac{\partial N}{\partial x}=\frac{x^2-y^2-2xy}{(x^+y^2)^2} $$ So, that would suggest conservative vector field, right? My question is how do I continue? Do I find potential function? $$ f(x,y)=\frac{1}{2}\ln(x^2+y^2)+\arctan(\frac{x}{y})+C$$ What next?
You have calculated that $$\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = -2a^2\pi.$$ (Remember to write the path over which you integrate as the "lower bound" of the integral! Also, the path you've written is not a circle unless $a = 1$. It's worth graphing for your self.) But you have also calculated that the vector field $$\mathbf{F} = \left(\frac{x+y}{x^2+y^2},\frac{-(x-y)}{x^2+y^2}\right)$$ is irrotational (it's curl is 0). If the vector field were conservative, then we ought to have $$\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = 0$$ without even needing to calculate a potential function. This is a basic property of conservative vector fields. So either your direct calculation of the integral is wrong or the field is irrotational but not conservative.
A vector field can be irrotational but not conservative if it is differentiable over a region which is not simply-connected. In the plane, a region is simply-connected if it has no holes. But in fact, the vector field $\mathbf{F}$ is not continuous at the point $(0,0)$ (why?), so it is certainly not differentiable. The path $\gamma$ circles the point $(0,0)$, so any region containing $\gamma$ contains $(0,0)$. We can conclude that $\mathbf{F}$ is not conservative in any region containing $\gamma$.