What is the integral of \sqrt[3]{(x^2-4x+4)^4} ?

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

integral of \sqrt[3]{(x^2-4x+4)^4}

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

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

Solution steps

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

3 (x^{2} - 4 x + 4)^{4} = (x^{2} - 4 x + 4)^{\frac{4}{3}},

assuming

(x^{2} - 4 x + 4) \geq 0

= \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

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