What is the integral of (3y-1)sqrt(3y^2-2y) ?

The integral of (3y-1)sqrt(3y^2-2y) is 1/(9sqrt(3))(9y^2-6y)^{3/2}+C

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integral of (3y-1)sqrt(3y^2-2y)

\int (3 y - 1) 3 y^{2} - 2 y d y

\frac{1}{9 3} (9 y^{2} - 6 y)^{\frac{3}{2}} + C

Solution steps

\int (3 y - 1) 3 y^{2} - 2 y d y

Apply u-substitution

: \int \frac{u u ^{2} - 1}{3 3} d u

= \int \frac{u u ^{2} - 1}{3 3} d u

Take the constant out:

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

= \frac{1}{3 3} \cdot \int u u^{2} - 1 d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{3 3} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{3 3} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{3 3} \cdot \frac{( ( 3 y - 1 ) ^{2} - 1 ) ^{3}}{3}

Simplify

\frac{1}{3 3} \cdot \frac{( ( 3 y - 1 ) ^{2} - 1 ) ^{3}}{3} : \frac{1}{9 3} (9 y^{2} - 6 y)^{\frac{3}{2}}

= \frac{1}{9 3} (9 y^{2} - 6 y)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9 3} (9 y^{2} - 6 y)^{\frac{3}{2}} + C

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What is the integral of (3y-1)sqrt(3y^2-2y) ?
The integral of (3y-1)sqrt(3y^2-2y) is 1/(9sqrt(3))(9y^2-6y)^{3/2}+C