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

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

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integral from 0 to 2 of 1/((x-1)^{1/3)}

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

0

Solution steps

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

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

Simplify

- \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + \frac{3}{2} : 0

= 0

\frac{d}{d y} (\frac{y ^{3}}{21} + \frac{7}{4 y})

f (x) = 8 x + ln (x^{2})

\int (5 x^{3} + 6)^{4} x^{2} d x

y^{^{'}} - y = 5 t e^{2 t}, y (0) = 1

n = 0 \sum \infty (1.075)^{n}

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