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.
2 Answers
$\begingroup$By the quotient rule $$\frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right) = \frac{x^2 - y^2}{(x^2 + y^2)^2}.$$
Therefore
$$\int_0^x \frac{x^2 - y^2}{(x^2 + y^2)^2}\, dy = \int_0^x \frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right)\, dy = \frac{y}{x^2 + y^2}\bigg|_{y = 0}^{y = x} = \frac{1}{2x}.$$
$\endgroup$ 2 $\begingroup$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*}
$\endgroup$
