What is the integral of 1/(x^2yz) ?

The integral of 1/(x^2yz) is -1/(yzx)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2yz)

\int \frac{1}{x ^{2} yz} d x

- \frac{1}{yz x} + C

Solution steps

\int \frac{1}{x ^{2} yz} d x

Take the constant out:

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

= \frac{1}{yz} \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

= \frac{1}{yz} (- \frac{1}{x})

Simplify

\frac{1}{yz} (- \frac{1}{x}) : - \frac{1}{yz x}

= - \frac{1}{yz x}

Add a constant to the solution

= - \frac{1}{yz x} + C