What is the integral of 1/(16x^4) ?

The integral of 1/(16x^4) is -1/(48x^3)+C

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integral of 1/(16x^4)

\int \frac{1}{16 x ^{4}} d x

- \frac{1}{48 x ^{3}} + C

Solution steps

\int \frac{1}{16 x ^{4}} 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 ^{4}} d x

Apply the Power Rule

: - \frac{1}{3 x ^{3}}

= \frac{1}{16} (- \frac{1}{3 x ^{3}})

Simplify

\frac{1}{16} (- \frac{1}{3 x ^{3}}) : - \frac{1}{48 x ^{3}}

= - \frac{1}{48 x ^{3}}

Add a constant to the solution

= - \frac{1}{48 x ^{3}} + C

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