What is the integral from 2 to \sqrt[3]{3 of} 1/x ?

The integral from 2 to \sqrt[3]{3 of} 1/x is -ln(2)+1/3 ln(3)

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integral from 2 to \sqrt[3]{3 of} 1/x

- ln (2) + \frac{1}{3} ln (3)

Decimal Notation

- 0.32694 \dots

Solution steps

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

Compute the boundaries:

ln (2) - \frac{1}{3} ln (3)

= - (ln (2) - \frac{1}{3} ln (3))

Simplify

= - ln (2) + \frac{1}{3} ln (3)

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What is the integral from 2 to \sqrt[3]{3 of} 1/x ?
The integral from 2 to \sqrt[3]{3 of} 1/x is -ln(2)+1/3 ln(3)