What is the integral from 8 to 27 of 1/(x-x^{1/3)} ?

The integral from 8 to 27 of 1/(x-x^{1/3)} is 3/2 (3ln(2)-ln(3))

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integral from 8 to 27 of 1/(x-x^{1/3)}

\int_{8}^{27} \frac{1}{x - x ^{\frac{1}{3}}} d x

\frac{3}{2} (3 ln (2) - ln (3))

Decimal Notation

1.47124 \dots

Solution steps

\int_{8}^{27} \frac{1}{x - x ^{\frac{1}{3}}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int_{3}^{8} \frac{1}{2 v} d v

= 3 \cdot \int_{3}^{8} \frac{1}{2 v} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{3}^{8} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 3 \cdot \frac{1}{2} [ln ∣ v ∣]_{3}^{8}

Simplify

3 \cdot \frac{1}{2} [ln ∣ v ∣]_{3}^{8} : \frac{3}{2} [ln ∣ v ∣]_{3}^{8}

= \frac{3}{2} [ln ∣ v ∣]_{3}^{8}

Compute the boundaries:

3 ln (2) - ln (3)

= \frac{3}{2} (3 ln (2) - ln (3))

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What is the integral from 8 to 27 of 1/(x-x^{1/3)} ?
The integral from 8 to 27 of 1/(x-x^{1/3)} is 3/2 (3ln(2)-ln(3))