What is the integral of (-9/(x^4)) ?

The integral of (-9/(x^4)) is 3/(x^3)+C

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integral of (-9/(x^4))

\int (- \frac{9}{x ^{4}}) d x

\frac{3}{x ^{3}} + C

Solution steps

\int - \frac{9}{x ^{4}} d x

Take the constant out:

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

= - 9 \cdot \int \frac{1}{x ^{4}} d x

Apply the Power Rule

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

= - 9 (- \frac{1}{3 x ^{3}})

Simplify

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

= \frac{3}{x ^{3}}

Add a constant to the solution

= \frac{3}{x ^{3}} + C

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