What is the integral of (x^2-4x+4)^{4/3}(3x-6) ?

The integral of (x^2-4x+4)^{4/3}(3x-6) is 1/(126*9^{1/3)}(3x-6)^{14/3}+C

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integral of (x^2-4x+4)^{4/3}(3x-6)

\int (x^{2} - 4 x + 4)^{\frac{4}{3}} (3 x - 6) d x

\frac{1}{126 \cdot 9 ^{\frac{1}{3}}} (3 x - 6)^{\frac{14}{3}} + C

Solution steps

\int (x^{2} - 4 x + 4)^{\frac{4}{3}} (3 x - 6) d x

Apply u-substitution

: \int \frac{u ^{\frac{11}{3}}}{27 \cdot 9 ^{\frac{1}{3}}} d u

= \int \frac{u ^{\frac{11}{3}}}{27 \cdot 9 ^{\frac{1}{3}}} d u

Take the constant out:

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

= \frac{1}{27 \cdot 9 ^{\frac{1}{3}}} \cdot \int u^{\frac{11}{3}} d u

Apply the Power Rule

: \frac{3}{14} u^{\frac{14}{3}}

= \frac{1}{27 \cdot 9 ^{\frac{1}{3}}} \cdot \frac{3}{14} u^{\frac{14}{3}}

Substitute back

u = 3 x - 6

= \frac{1}{27 \cdot 9 ^{\frac{1}{3}}} \cdot \frac{3}{14} (3 x - 6)^{\frac{14}{3}}

Simplify

\frac{1}{27 \cdot 9 ^{\frac{1}{3}}} \cdot \frac{3}{14} (3 x - 6)^{\frac{14}{3}} : \frac{1}{126 \cdot 9 ^{\frac{1}{3}}} (3 x - 6)^{\frac{14}{3}}

= \frac{1}{126 \cdot 9 ^{\frac{1}{3}}} (3 x - 6)^{\frac{14}{3}}

Add a constant to the solution

= \frac{1}{126 \cdot 9 ^{\frac{1}{3}}} (3 x - 6)^{\frac{14}{3}} + C

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What is the integral of (x^2-4x+4)^{4/3}(3x-6) ?
The integral of (x^2-4x+4)^{4/3}(3x-6) is 1/(126*9^{1/3)}(3x-6)^{14/3}+C