What is the integral from 0 to 4 of 1/((x-1)^{1/3)} ?

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

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integral from 0 to 4 of 1/((x-1)^{1/3)}

\int_{0}^{4} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

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

Decimal Notation

1.62012 \dots

Solution steps

\int_{0}^{4} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

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

= \int_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x + \int_{1}^{4} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

\int_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x = - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2}

\int_{1}^{4} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x = \frac{3 \cdot 3 ^{\frac{2}{3}}}{2}

= - \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}

\int_{1}^{\infty} \frac{2}{3 x + 4} d x

\int_{- 2}^{1} x^{2} + x + 2 d x

\int_{- 1}^{1} \frac{5}{1 + 25 x ^{2}} d x

\int_{y}^{y^{2}} (x + 2 y) d x

\int_{0}^{3} (1 + 2 x)^{2} d x

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