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

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

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

= - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + 6

Simplify

- \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + 6 : \frac{9}{2}

= \frac{9}{2}

f (x) = x^{3} - 2 x^{2} - 5 x + 6, at x = 2

\int_{0}^{3} 15 8 x + 1 d x

\int g e^{- c t} d t

\frac{d}{d x} (a x sin (4 x) + b x cos (4 x))

f^{^{'}} (t) - f (t) = 2 t

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