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

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

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

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

- arctan (\frac{x}{y}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= - y \frac{1}{y} arctan (u)

Substitute back

u = \frac{x}{y}

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

Simplify

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

= - arctan (\frac{x}{y})

Add a constant to the solution

= - arctan (\frac{x}{y}) + C

\int ln (2 + x^{2}) d x

\int_{0}^{8} \frac{1}{64 + t ^{2}} d t

\frac{d}{d x} (1 + 4 x)

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

y = x^{4} + 6 x^{2} - x

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