The integral is path-independent if we can find a scalar function f such that grad(f ) = A. This requires
[tex]\dfrac{\partial f}{\partial x}=2x-y[/tex]
[tex]\dfrac{\partial f}{\partial y}=x+y[/tex]
Take the first PDE and integrate both sides with respect to x to get
[tex]f(x,y)=x^2-xy+g(y)[/tex]
where g is assumed to be a function of y alone. Differentiating this with respect to x gives
[tex]\dfrac{\partial f}{\partial y}=-x+\dfrac{\mathrm dg}{\mathrm dy}=x+y\implies\dfrac{\mathrm dg}{\mathrm dy}=2x+y[/tex]
which would mean g is *not* a function of only y, but also x, contradicting our assumption. So the integral is path-dependent.
Parameterize the unit circle (call it C) by the vector function,
[tex]\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j[/tex]
with t between 0 and 2π.
Note that this parameterization takes C to have counter-clockwise orientation; if we compute the line integral of A over C, we can multiply the result by -1 to get the value of the integral in the opposite, clockwise direction.
Then
[tex]\mathrm d\mathbf r=-\sin t\,\mathbf i+\cos t\,\mathbf j[/tex]
and the (counter-clockwise) integral over C is
[tex]\displaystyle\int_C\mathbf A\cdot\mathrm d\mathbf r[/tex]
[tex]\displaystyle=\int_0^{2\pi}((2\cos t-\sin t)\,\mathbf i+(\cos t+\sin t)\,\mathbf j)\cdot(-\sin t\,\mathbf i+\cos t\,\mathbf j)\,\mathrm dt[/tex]
[tex]\displaystyle=\int_0^{2\pi}1-\sin t\cos t\,\mathrm dt=2\pi[/tex]
and so the integral in the direction we want is -2π.
By the way, that the integral doesn't have a value of 0 is more evidence of the fact that the integral is path-dependent.
