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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int_{1}^{1 - x} - \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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

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