What is the integral from 0 to 9 of 1/((x-1)^{2/3)} ?

The integral from 0 to 9 of 1/((x-1)^{2/3)} is 9

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integral from 0 to 9 of 1/((x-1)^{2/3)}

\int_{0}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

9

Solution steps

\int_{0}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{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{2}{3}}} d x + \int_{1}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

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

\int_{1}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x = 6

= - 3 (- 1)^{\frac{1}{3}} + 6

Simplify

- 3 (- 1)^{\frac{1}{3}} + 6 : 9

= 9

\frac{\partial}{\partial x} (e^{- 2 y} sin (3 x))

\int t ec (2 t) d t

\frac{d}{d x} ((x + 1) \cdot (x^{2} - 1))

\frac{x ^{2} - 3}{x - 2}

\int_{- 1}^{6} (2 x + 6) - (x^{2} - 3 x) d x

What is the integral from 0 to 9 of 1/((x-1)^{2/3)} ?
The integral from 0 to 9 of 1/((x-1)^{2/3)} is 9