What is the integral from 1 to 3 of 2/((x-2)^{8/3)} ?

The integral from 1 to 3 of 2/((x-2)^{8/3)} is diverges

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integral from 1 to 3 of 2/((x-2)^{8/3)}

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 2

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_{1}^{2} \frac{2}{( x - 2 ) ^{\frac{8}{3}}} d x + \int_{2}^{3} \frac{2}{( x - 2 ) ^{\frac{8}{3}}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

2 x^{2} - 5

\int_{- 3}^{3} x^{3} d x

\frac{d}{d x} (ln (1 + 3 x) + x ln (x))

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

\frac{d}{d x} (\frac{1}{( 3 x ^{2} + x + 2 ) ^{5}})

What is the integral from 1 to 3 of 2/((x-2)^{8/3)} ?
The integral from 1 to 3 of 2/((x-2)^{8/3)} is diverges