What is the integral from 0 to 9 of 8/((x-1)^{1/3)} ?

The integral from 0 to 9 of 8/((x-1)^{1/3)} is 36

Popular >

integral from 0 to 9 of 8/((x-1)^{1/3)}

\int_{0}^{9} \frac{8}{( x - 1 ) ^{\frac{1}{3}}} d x

36

Solution steps

\int_{0}^{9} \frac{8}{( 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{8}{( x - 1 ) ^{\frac{1}{3}}} d x + \int_{1}^{9} \frac{8}{( x - 1 ) ^{\frac{1}{3}}} d x

\int_{0}^{1} \frac{8}{( x - 1 ) ^{\frac{1}{3}}} d x = - 12 (- 1)^{\frac{2}{3}}

\int_{1}^{9} \frac{8}{( x - 1 ) ^{\frac{1}{3}}} d x = 48

= - 12 (- 1)^{\frac{2}{3}} + 48

Simplify

- 12 (- 1)^{\frac{2}{3}} + 48 : 36

= 36

\frac{\partial}{\partial x} (9 x \cdot cos (x) \cdot cos (y))

\frac{d}{d x} (- 2 cos (x) - x)

\int sin (a - x) d x

x \to 2 lim (\frac{( x + 3 )}{x + 4})

\frac{d}{d t} (t sin (π t))

What is the integral from 0 to 9 of 8/((x-1)^{1/3)} ?
The integral from 0 to 9 of 8/((x-1)^{1/3)} is 36