What is the integral from 4 to 5 of 1/((x-4)^{4/3)} ?

The integral from 4 to 5 of 1/((x-4)^{4/3)} is diverges

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integral from 4 to 5 of 1/((x-4)^{4/3)}

\int_{4}^{5} \frac{1}{( x - 4 ) ^{\frac{4}{3}}} d x

diverges

Solution steps

\int_{4}^{5} \frac{1}{( x - 4 ) ^{\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

f (x) = e^{x + 2} + 1

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

x \to 0 lim (ln ∣ x ∣)

\frac{d}{d x} (\frac{x + 3}{2})

y^{^{''}} + 2 y^{^{'}} + 1 = 0

What is the integral from 4 to 5 of 1/((x-4)^{4/3)} ?
The integral from 4 to 5 of 1/((x-4)^{4/3)} is diverges