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

The integral of (2y)/((x^2+y^2)^2) is -1/(x^2+y^2)+C

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

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

- \frac{1}{x ^{2} + y ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x}{7 + x})

\frac{d}{d x} (\frac{8}{x ^{9}})

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{4})

\frac{d}{d x} (7 x + 9)

f (t) = (1 + t) (2 t^{2} - 7)

What is the integral of (2y)/((x^2+y^2)^2) ?
The integral of (2y)/((x^2+y^2)^2) is -1/(x^2+y^2)+C