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 approach is
$\int \frac{y^2-x^2}{(x^2+y^2)^2} dy = \int\frac{y^2+x^2}{(x^2+y^2)^2} dy - \int \frac{2x^2}{(x^2+y^2)^2} dy = \int\frac{1}{x^2+y^2} dy - 2x^2\int \frac{1}{(x^2+y^2)^2} dy$
Once in this form, simple variable substitution for the first one and integration by parts or by $y = tan(u)$ substitution should be very straight forward.
