What is the integral from 0 to 9 of 6/((x-1)^{1/3)} ?

The integral from 0 to 9 of 6/((x-1)^{1/3)} is 27

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integral from 0 to 9 of 6/((x-1)^{1/3)}

\int_{0}^{9} \frac{6}{( x - 1 ) ^{\frac{1}{3}}} d x

27

Solution steps

\int_{0}^{9} \frac{6}{( 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{6}{( x - 1 ) ^{\frac{1}{3}}} d x + \int_{1}^{9} \frac{6}{( x - 1 ) ^{\frac{1}{3}}} d x

\int_{0}^{1} \frac{6}{( x - 1 ) ^{\frac{1}{3}}} d x = - 9 (- 1)^{\frac{2}{3}}

\int_{1}^{9} \frac{6}{( x - 1 ) ^{\frac{1}{3}}} d x = 36

= - 9 (- 1)^{\frac{2}{3}} + 36

Simplify

- 9 (- 1)^{\frac{2}{3}} + 36 : 27

= 27

10 x

\int \frac{2}{3} d x

y = x^{2} + 7, (- 4, 23)

\frac{f ( x ) g ( x )}{f ( x ) - g ( x )}

y^{^{'}} = 5 y + a

What is the integral from 0 to 9 of 6/((x-1)^{1/3)} ?
The integral from 0 to 9 of 6/((x-1)^{1/3)} is 27