What is the integral of 1/(18x^5) ?

The integral of 1/(18x^5) is -1/(72x^4)+C

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integral of 1/(18x^5)

\int \frac{1}{18 x ^{5}} d x

- \frac{1}{72 x ^{4}} + C

Solution steps

\int \frac{1}{18 x ^{5}} d x

Take the constant out:

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

= \frac{1}{18} \cdot \int \frac{1}{x ^{5}} d x

Apply the Power Rule

: - \frac{1}{4 x ^{4}}

= \frac{1}{18} (- \frac{1}{4 x ^{4}})

Simplify

\frac{1}{18} (- \frac{1}{4 x ^{4}}) : - \frac{1}{72 x ^{4}}

= - \frac{1}{72 x ^{4}}

Add a constant to the solution

= - \frac{1}{72 x ^{4}} + C

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