How to integrate $(x^2 - y^2) / (x^2 + y^2)^2$ How do I integrate 
$$\int \int \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dx dy?$$
The WolframAlpha page gives 
$$
c_1 + c_2 + \tan^{-1}(x/y).
$$
And I kind of specifically need
$$
\int_{0}^{x} \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dy.
$$
Note


*

*I want to know integration technique to solve this without using $F' = f$.

*For the double integral above, what I'm interested is Lebesgue integral, but I guess what Wolfram gave is in the Riemann sense.


The main integral
$$I = \int\frac{x^2-y^2}{(x^2+y^2)^2}\,dx$$
can be solved using a tan substitution.
\begin{align*}
x &= y\tan\theta \\
dx &= y\sec^2\theta\,d\theta
\end{align*}
So that
\begin{align*}
I &= \int \frac{y^2(\tan^2\theta -1)}{y^4\sec^4\theta}\,y\sec^2\theta\,d\theta \\
&= \frac{1}{y}\int \frac{\tan^2\theta-1}{\sec^2\theta}\,d\theta \\
&= \frac{1}{y}\int \sin^2\theta - \cos^2\theta\,d\theta \\
&= - \frac{1}{y}\int \cos 2\theta\,d\theta \\
&= - \frac{1}{y}\frac{1}{2}\sin 2\theta + h(y) \\
&= - \frac{1}{2y} \frac{2\tan\theta}{1+\tan^2\theta}  + h(y) \\
&= - \frac{1}{2y} \frac{2 (y/x)}{1+(y/x)^2} + h(y) \\
&= - \frac{x}{x^2+y^2} + h(y).
\end{align*}
