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

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

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

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

- x ln x^{2} + y^{2} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 x \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Simplify

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

= - x ln x^{2} + y^{2}

Add a constant to the solution

= - x ln x^{2} + y^{2} + C

\frac{\partial}{\partial x} ((1 + x) sin (y))

\frac{d}{d x} (ln (a x a + x))

\int \frac{1}{( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

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

t \to 0 lim (\frac{6}{1 + t})

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