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

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

Apply the Power Rule

= \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

\int tan^{2} (\frac{2}{x}) \frac{1}{x ^{2}} d x

\int e^{i θ} d θ

\int \frac{3}{2 + x ^{3}} d x

\int sin^{6} (\frac{x}{3}) cos (\frac{x}{3}) d x

\int e^{x} \cdot e^{e^{x}} d 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