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

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

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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}{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

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