Line integral - should I parametrize the square? I have the following $1-\text{form}$ defined: $$\omega = \displaystyle\frac{2xy}{(1-x^2)^2+y^2}\mathrm{dx}+\displaystyle\frac{1-x^2}{(1-x^2)^2+y^2}\mathrm{dy}$$ I'd like to find $\displaystyle\oint_{\gamma}\omega$ where $\gamma$ is the square defined by the points $(2,-1),(2,1),(0,1),(0,-1)$ counter clockwise.
Since the square is a closed path, if $\omega$ is exact then $\displaystyle\oint_\gamma \omega =0$, right?
This is something I have proved using $$\int\omega=y\int\frac{2x}{(1-x^2)^2}+y^2 dx  \\ \text{change variables: }u= (1-x^2)\implies du=-2x \\ =-y\int \frac{1}{(u/y)^2+1}du \\ \text{2nd change of variables: } z=u/y \implies dz=du/y \\ =-\int \frac{1}{z^2+1}dz = - \arctan(z)\\=-\arctan\left(\frac{1-x^2}{y}\right) + C(y)$$
And it seems to be enough if I set $C(y)=0$ since 
$$\frac{\partial}{\partial y} \left(-\arctan\frac{1-x^2}{y}\right)=-\frac{1}{\frac{(1-x^2)^2}{y^2}+1}\frac{\partial}{\partial y}\left(\frac{1-x^2}{y}\right) \\ =-\frac{y^2}{(1-x^2)^2+y^2}\left(-\frac{1-x^2}{y^2}\right) = \frac{1-x^2}{(1-x^2)^2+y^2}$$
Is this enough to show that $\displaystyle\oint_\gamma \omega =0$ ?

Have you thought of Green's theorem? It's quite simple to solve if you use Green's theorem.
