What is the integral from 9 to 10 of 1/((x-9)^{4/3)} ?

The integral from 9 to 10 of 1/((x-9)^{4/3)} is diverges

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integral from 9 to 10 of 1/((x-9)^{4/3)}

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{4}{3}}} d x

diverges

Solution steps

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{4}{3}}} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{\frac{4}{3}}} d u

= \int_{0}^{1} \frac{1}{u ^{\frac{4}{3}}} d u

Apply the Power Rule

: [- \frac{3}{u ^{\frac{1}{3}}}]_{0}^{1}

= [- \frac{3}{u ^{\frac{1}{3}}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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\frac{d}{d x} (arctan (x^{5}))

\frac{\partial}{\partial y} (y^{2 x})

What is the integral from 9 to 10 of 1/((x-9)^{4/3)} ?
The integral from 9 to 10 of 1/((x-9)^{4/3)} is diverges