What is the integral of-(9x)^{1/3} ?

The integral of-(9x)^{1/3} is -9^{1/3}*3/4 x^{4/3}+C

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integral of-(9x)^{1/3}

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

- 9^{\frac{1}{3}} \cdot \frac{3}{4} x^{\frac{4}{3}} + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{4} x^{\frac{4}{3}}

= - 9^{\frac{1}{3}} \cdot \frac{3}{4} x^{\frac{4}{3}}

Add a constant to the solution

= - 9^{\frac{1}{3}} \cdot \frac{3}{4} x^{\frac{4}{3}} + C

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What is the integral of-(9x)^{1/3} ?
The integral of-(9x)^{1/3} is -9^{1/3}*3/4 x^{4/3}+C