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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

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

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

3 x^{2} + \frac{8 x}{3} - 2

2 y^{^{'}} + y = 8 t^{2}

\frac{\partial}{\partial y} (4 - x^{2} - 2 y)

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