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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Apply the Power Rule

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

Compute the boundaries:

diverges

= diverges

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