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

The integral from 1 to 9 of 9/((x-1)^{1/3)} is 54

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

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

54

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 9 \cdot \int_{0}^{8} \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

: [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{8}

= 9 [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{8}

Compute the boundaries:

6

= 9 \cdot 6

Simplify

= 54

\int_{1}^{\infty} \frac{1}{n ^{5}}

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What is the integral from 1 to 9 of 9/((x-1)^{1/3)} ?
The integral from 1 to 9 of 9/((x-1)^{1/3)} is 54