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

The integral of (1/3 x-9)^{2/3} is 9/5 (1/3 (x-27))^{5/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{9}{5} (\frac{1}{3} (x - 27))^{\frac{5}{3}} + C

Solution steps

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

Apply u-substitution

: \int u^{\frac{2}{3}} \cdot 3 d u

= \int u^{\frac{2}{3}} \cdot 3 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int u^{\frac{2}{3}} d u

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= 3 \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = \frac{1}{3} x - 9

= 3 \cdot \frac{3}{5} (\frac{1}{3} x - 9)^{\frac{5}{3}}

Simplify

3 \cdot \frac{3}{5} (\frac{1}{3} x - 9)^{\frac{5}{3}} : \frac{9}{5} (\frac{1}{3} (x - 27))^{\frac{5}{3}}

= \frac{9}{5} (\frac{1}{3} (x - 27))^{\frac{5}{3}}

Add a constant to the solution

= \frac{9}{5} (\frac{1}{3} (x - 27))^{\frac{5}{3}} + C

\int - 25 sin (5 x) d x

\frac{d y}{d t} = 45 t^{8}

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

y^{^{''}} + y^{^{'}} + \frac{1}{4} y = 0

What is the integral of (1/3 x-9)^{2/3} ?
The integral of (1/3 x-9)^{2/3} is 9/5 (1/3 (x-27))^{5/3}+C