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I'm self-studying a Cramster solution and I came across this integral and I don't know what they've done with it. Help would be appreciated.

$$\int \frac{y^2 - x^2}{(x^2 + y^2)^2} ~dy.$$

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Is x constant wrt y? –  lab bhattacharjee Jul 17 '12 at 7:38

2 Answers 2

up vote 3 down vote accepted

In an integral $dy$, $x$ is a constant. Rewirite this as: $$\int\frac{y^2-x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{y^2+x^2-2x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{dy}{x^2+y^2}-2x^2\int\frac{dy}{\left(x^2+y^2\right)^2}$$ Can you continue from here?

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This is the same as the "partial fractions" method, correct? I think I can take it from here. Thanks. –  Joebevo Jul 17 '12 at 4:17
    
Yes, this is a simple example of the partial fractions method. –  Dennis Gulko Jul 17 '12 at 5:19

If x is constant wrt y,

let $y=x. tan(z)$, then $dy=xsec^2zdz $

Then $$\int \frac{y^2 - x^2}{(x^2 + y^2)^2} ~dy$$ becomes

$$\frac{-1}{x}\int cos(2z) ~dz$$

=$\frac{-sin(2z)}{2x} + C$, where C is an undetermined constant.

=$\frac{-tan(z)}{x(1+tan^2z)} +C$

=$\frac{-y}{x^2+y^2}+ C$

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