What is the integral from 1 to 9 of 1/((x-1)^{2/3)} ?

The integral from 1 to 9 of 1/((x-1)^{2/3)} is 6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 9 of 1/((x-1)^{2/3)}

\int_{1}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

6

Solution steps

\int_{1}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

Apply u-substitution

: \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

= \int_{0}^{8} \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

: [3 u^{\frac{1}{3}}]_{0}^{8}

= [3 u^{\frac{1}{3}}]_{0}^{8}

Compute the boundaries:

6

= 6

\int_{0}^{t} 3 t^{3} d t

\int_{- 3}^{2} - x^{2} d x

\int_{1}^{x} t^{2} - 1 d t

\int_{0}^{2} e^{x} x^{3} d x

\int_{0}^{4} ln (1 + x) d x

What is the integral from 1 to 9 of 1/((x-1)^{2/3)} ?
The integral from 1 to 9 of 1/((x-1)^{2/3)} is 6