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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= x \frac{1}{x ^{2}} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{2}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{2}{( u ^{2} + 1 ) ^{2}} d u = arctan (u) + \frac{u}{1 + u ^{2}}

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

Substitute back

u = \frac{y}{x}

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

Simplify

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

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

Add a constant to the solution

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

n = 2 \sum \infty \frac{6}{( n ^{2} - 1 )}

\frac{d}{d x} (\frac{9 - x ^{2}}{4 x})

\frac{\partial}{\partial x} (\frac{r t}{r + t})

\frac{d}{d x} (x^{\frac{2}{3}} (x + 4)^{\frac{1}{2}})

x^{2} y = (x^{3} + y^{3}) y^{'}

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