What is the integral of (-9)/(\sqrt[3]{x^5)} ?

The integral of (-9)/(\sqrt[3]{x^5)} is (27)/(2x^{2/3)}+C

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integral of (-9)/(\sqrt[3]{x^5)}

\frac{27}{2 x ^{\frac{2}{3}}} + C

Solution steps

Take the constant out:

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

assuming

x \geq 0

= - 9 \cdot \int \frac{1}{x ^{\frac{5}{3}}} d x

Apply the Power Rule

: - \frac{3}{2 x ^{\frac{2}{3}}}

= - 9 (- \frac{3}{2 x ^{\frac{2}{3}}})

Simplify

- 9 (- \frac{3}{2 x ^{\frac{2}{3}}}) : \frac{27}{2 x ^{\frac{2}{3}}}

= \frac{27}{2 x ^{\frac{2}{3}}}

Add a constant to the solution

= \frac{27}{2 x ^{\frac{2}{3}}} + C

\frac{d}{d x} (1 - x^{3})

x \to 0 lim (\frac{sec ( x ) tan ( x )}{x})

\frac{d y}{d x} = \frac{e ^{y} + e ^{- y}}{e ^{x} - e ^{- x}}

\int (e^{2 x} + sin (2 x)) d x

\frac{d}{d x} (sin (x) + \frac{8}{3} cot (x))

What is the integral of (-9)/(\sqrt[3]{x^5)} ?
The integral of (-9)/(\sqrt[3]{x^5)} is (27)/(2x^{2/3)}+C