What is the integral of (x^2-4x+4)^{4/3} ?

The integral of (x^2-4x+4)^{4/3} is 3/11 (x-2)^{11/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-4x+4)^{4/3}

\int (x^{2} - 4 x + 4)^{\frac{4}{3}} d x

\frac{3}{11} (x - 2)^{\frac{11}{3}} + C

Solution steps

\int (x^{2} - 4 x + 4)^{\frac{4}{3}} d x

Complete the square

x^{2} - 4 x + 4 : (x - 2)^{2}

= \int ((x - 2)^{2})^{\frac{4}{3}} d x

Apply u-substitution

: \int u^{\frac{8}{3}} d u

= \int u^{\frac{8}{3}} d u

Apply the Power Rule

: \frac{3}{11} u^{\frac{11}{3}}

= \frac{3}{11} u^{\frac{11}{3}}

Substitute back

u = x - 2

= \frac{3}{11} (x - 2)^{\frac{11}{3}}

Add a constant to the solution

= \frac{3}{11} (x - 2)^{\frac{11}{3}} + C

\frac{d y}{d x} = \frac{5 x}{6 y}

\int (2 x^{2} + 5 x) e^{3 x} d x

\frac{e ^{- a s}}{s ^{2}}

3 + 4 x^{2} - 5 x^{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 x}

What is the integral of (x^2-4x+4)^{4/3} ?
The integral of (x^2-4x+4)^{4/3} is 3/11 (x-2)^{11/3}+C