What is the integral of 1/2 y^{-2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/2 y^{-2}

\int \frac{1}{2} y^{- 2} d y

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

Solution steps

\int \frac{1}{2} y^{- 2} d y

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{y}

= \frac{1}{2} (- \frac{1}{y})

Simplify

\frac{1}{2} (- \frac{1}{y}) : - \frac{1}{2 y}

= - \frac{1}{2 y}

Add a constant to the solution

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

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

\int \frac{x - 5}{x ^{2} ( x - 1 )} d x

\int \frac{x - 42}{x ^{2} ( x + 3 )} d x

\int - 1280 (1 + 8 x)^{- 3} d x

\int 4 x + 3 y d x