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

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

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

\int_{0}^{9} \frac{1}{∣ x - 1 ∣ ^{\frac{2}{3}}} d x

9

Solution steps

\int_{0}^{9} \frac{1}{∣ x - 1 ∣ ^{\frac{2}{3}}} d x

Eliminate Absolutes

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

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

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

= 3 + 6

Simplify

= 9

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