What is the integral of (16)/(x^2) ?

The integral of (16)/(x^2) is -(16)/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16)/(x^2)

\int \frac{16}{x ^{2}} d x

- \frac{16}{x} + C

Solution steps

\int \frac{16}{x ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

= 16 (- \frac{1}{x})

Simplify

16 (- \frac{1}{x}) : - \frac{16}{x}

= - \frac{16}{x}

Add a constant to the solution

= - \frac{16}{x} + C

\int_{0}^{4} x^{2} 16 - x^{2} d x

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

n = 0 \sum \infty \frac{1}{2 ^{2 n}}

\int \frac{1}{6} e^{6 x} d x