What is the integral from 1/2 to 5 of 1/((2x-1)^{1/3)} ?

The integral from 1/2 to 5 of 1/((2x-1)^{1/3)} is (3*9^{2/3})/4

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integral from 1/2 to 5 of 1/((2x-1)^{1/3)}

\int_{\frac{1}{2}}^{5} \frac{1}{( 2 x - 1 ) ^{\frac{1}{3}}} d x

\frac{3 \cdot 9 ^{\frac{2}{3}}}{4}

Decimal Notation

3.24506 \dots

Solution steps

\int_{\frac{1}{2}}^{5} \frac{1}{( 2 x - 1 ) ^{\frac{1}{3}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int_{\frac{1}{2}}^{5} (2 x - 1)^{- \frac{1}{3}} d x

Apply u-substitution

: \int_{0}^{9} \frac{1}{2 u ^{\frac{1}{3}}} d u

= \int_{0}^{9} \frac{1}{2 u ^{\frac{1}{3}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{9} \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

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

= \frac{1}{2} [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{9}

Compute the boundaries:

9^{\frac{2}{3}} \cdot \frac{3}{2}

= \frac{1}{2} \cdot 9^{\frac{2}{3}} \cdot \frac{3}{2}

Simplify

= \frac{3 \cdot 9 ^{\frac{2}{3}}}{4}

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What is the integral from 1/2 to 5 of 1/((2x-1)^{1/3)} ?
The integral from 1/2 to 5 of 1/((2x-1)^{1/3)} is (3*9^{2/3})/4