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

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

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

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

- 6 (- 1)^{\frac{1}{3}}

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{2}^{9} \frac{1}{( 1 - x ) ^{\frac{2}{3}}} d x

Apply u-substitution

: \int_{- 1}^{- 8} - \frac{1}{u ^{\frac{2}{3}}} d u

= 2 \cdot \int_{- 1}^{- 8} - \frac{1}{u ^{\frac{2}{3}}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 8}^{- 1} - \frac{1}{u ^{\frac{2}{3}}} d u)

Take the constant out:

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

= 2 (- (- \int_{- 8}^{- 1} \frac{1}{u ^{\frac{2}{3}}} d u))

Apply the Power Rule

: [3 u^{\frac{1}{3}}]_{- 8}^{- 1}

= 2 (- (- [3 u^{\frac{1}{3}}]_{- 8}^{- 1}))

Simplify

= 2 [3 u^{\frac{1}{3}}]_{- 8}^{- 1}

Compute the boundaries:

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

= 2 (- 3 (- 1)^{\frac{1}{3}})

Simplify

= - 6 (- 1)^{\frac{1}{3}}

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