What is the integral of (4x)^{-2} ?

The integral of (4x)^{-2} is -1/(16x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x)^{-2}

\int (4 x)^{- 2} d x

- \frac{1}{16 x} + C

Solution steps

\int (4 x)^{- 2} d x

Simplify

(4 x)^{- 2} : \frac{1}{16 x ^{2}}

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{x}

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

Simplify

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

= - \frac{1}{16 x}

Add a constant to the solution

= - \frac{1}{16 x} + C

\int \frac{e ^{x}}{( x + 1 ) ^{2}} d x

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

\int - \frac{6}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

\int \frac{sin ( 7 x )}{cos ( 7 x ) - 2} d x

\int 0.15 - 0.09 e^{0.006 x} d x