How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$?

Question

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.

Answer 1

$$ \int\frac{y^2-x^2}{(y^2+x^2)^2}dy $$ \begin{align} y & = x\tan\theta \\ dy & = x\sec^2\theta\,d\theta \\ y^2+x^2 & = x^2(\tan^2\theta+1) = x^2\sec^2\theta \\ y^2-x^2 & = x^2(\tan^2\theta-1) \\ & \phantom{=}\text{etc.} \end{align} Usually with $\displaystyle\int \Big(\cdots\cdots\text{something involving }(y^2+\text{constant}^2)\cdots\cdots\Big)\,dy$, this will work.

Answer 2

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

$\bigg(\displaystyle-\frac{y}{y^2+x^2}+\int\frac{1}{y^2+x^2}dy\bigg)-\int\frac{1}{y^2+x^2}dy=\frac{-y}{y^2+x^2}+g(x)$

Answer 3

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.