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

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

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

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

6

Solution steps

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

Apply u-substitution

= \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

= [3 u^{\frac{1}{3}}]_{0}^{8}

Compute the boundaries:

6

= 6

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