What is the integral of-(2xy)/((x^2+y^2)^2) ?

The integral of-(2xy)/((x^2+y^2)^2) is x/(x^2+y^2)+C

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integral of-(2xy)/((x^2+y^2)^2)

\int - \frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}} d y

\frac{x}{x ^{2} + y ^{2}} + C

Solution steps

\int - \frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 x \cdot \int \frac{y}{( x ^{2} + y ^{2} ) ^{2}} d y

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= - 2 x \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 x \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 2 x \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + y^{2}

= - 2 x \frac{1}{2} (- \frac{1}{x ^{2} + y ^{2}})

Simplify

- 2 x \frac{1}{2} (- \frac{1}{x ^{2} + y ^{2}}) : \frac{x}{x ^{2} + y ^{2}}

= \frac{x}{x ^{2} + y ^{2}}

Add a constant to the solution

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

x y^{^{'}} + 3 y = \frac{( 2 e ^{- 2 x} )}{x ^{2}}

x \to \infty lim (7 + \frac{1}{x})

x^{3} + y^{3} = 9 x y, (4, 2)

x \to - 1 lim (- 4 x^{9} - 2 x^{6} - x^{3})

What is the integral of-(2xy)/((x^2+y^2)^2) ?
The integral of-(2xy)/((x^2+y^2)^2) is x/(x^2+y^2)+C