What is the integral of 1/(16x^2) ?

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

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integral of 1/(16x^2)

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

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

Solution steps

\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