What is the integral from 0 to 4 of 1/(\sqrt[3]{x-1)} ?

The integral from 0 to 4 of 1/(\sqrt[3]{x-1)} is -3/2+(3*3^{2/3})/2

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integral from 0 to 4 of 1/(\sqrt[3]{x-1)}

- \frac{3}{2} + \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

Decimal Notation

1.62012 \dots

Solution steps

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

= - \frac{3}{2} + \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

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What is the integral from 0 to 4 of 1/(\sqrt[3]{x-1)} ?
The integral from 0 to 4 of 1/(\sqrt[3]{x-1)} is -3/2+(3*3^{2/3})/2