What is the integral from 0 to 6 of 1/((x-9)^{2/3)} ?

The integral from 0 to 6 of 1/((x-9)^{2/3)} is 3(-1)^{1/3}*3^{1/3}-3(-1)^{1/3}*9^{1/3}

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integral from 0 to 6 of 1/((x-9)^{2/3)}

\int_{0}^{6} \frac{1}{( x - 9 ) ^{\frac{2}{3}}} d x

3 (- 1)^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} - 3 (- 1)^{\frac{1}{3}} \cdot 9^{\frac{1}{3}}

Solution steps

\int_{0}^{6} \frac{1}{( x - 9 ) ^{\frac{2}{3}}} d x

Apply u-substitution

: \int_{- 9}^{- 3} \frac{1}{u ^{\frac{2}{3}}} d u

= \int_{- 9}^{- 3} \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

: [3 u^{\frac{1}{3}}]_{- 9}^{- 3}

= [3 u^{\frac{1}{3}}]_{- 9}^{- 3}

Compute the boundaries:

3 (- 1)^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} - 3 (- 1)^{\frac{1}{3}} \cdot 9^{\frac{1}{3}}

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

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What is the integral from 0 to 6 of 1/((x-9)^{2/3)} ?
The integral from 0 to 6 of 1/((x-9)^{2/3)} is 3(-1)^{1/3}*3^{1/3}-3(-1)^{1/3}*9^{1/3}