How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.

Another way to find this integral would be to use
$\displaystyle\int\frac{y^2-x^2}{(y^2+x^2)^2}dy=\int\bigg(\frac{2y^2}{(x^2+y^2)^2}-\frac{1}{y^2+x^2}\bigg)dy$;
Using integration by parts  for the first term with $u=y, dv=\frac{2y}{(x^2+y^2)^2}dy$ gives 
$\bigg(\displaystyle-\frac{y}{y^2+x^2}+\int\frac{1}{y^2+x^2}dy\bigg)-\int\frac{1}{y^2+x^2}dy=\frac{-y}{y^2+x^2}+g(x)$
