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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{u ^{\frac{1}{3}}} d u

= \int_{- 1}^{0} \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

: [\frac{3}{2} u^{\frac{2}{3}}]_{- 1}^{0}

= [\frac{3}{2} u^{\frac{2}{3}}]_{- 1}^{0}

Compute the boundaries:

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

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

Simplify

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

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