What is the integral of (3x-1)(6x^2-4x+1)^{1/3} ?

The integral of (3x-1)(6x^2-4x+1)^{1/3} is 3/16 (6x^2-4x+1)^{4/3}+C

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integral of (3x-1)(6x^2-4x+1)^{1/3}

\int (3 x - 1) (6 x^{2} - 4 x + 1)^{\frac{1}{3}} d x

\frac{3}{16} (6 x^{2} - 4 x + 1)^{\frac{4}{3}} + C

Solution steps

\int (3 x - 1) (6 x^{2} - 4 x + 1)^{\frac{1}{3}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{4} d u

= \int \frac{u ^{\frac{1}{3}}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{\frac{1}{3}} d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{4} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 6 x^{2} - 4 x + 1

= \frac{1}{4} \cdot \frac{3}{4} (6 x^{2} - 4 x + 1)^{\frac{4}{3}}

Simplify

\frac{1}{4} \cdot \frac{3}{4} (6 x^{2} - 4 x + 1)^{\frac{4}{3}} : \frac{3}{16} (6 x^{2} - 4 x + 1)^{\frac{4}{3}}

= \frac{3}{16} (6 x^{2} - 4 x + 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{16} (6 x^{2} - 4 x + 1)^{\frac{4}{3}} + C

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What is the integral of (3x-1)(6x^2-4x+1)^{1/3} ?
The integral of (3x-1)(6x^2-4x+1)^{1/3} is 3/16 (6x^2-4x+1)^{4/3}+C