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

The integral of y(3-3y)^2 is 3/2 y^4+(3y^4)/4-6y^3+9/2 y^2+C

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integral of y(3-3y)^2

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

\frac{3}{2} y^{4} + \frac{3 y ^{4}}{4} - 6 y^{3} + \frac{9}{2} y^{2} + C

Solution steps

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

Apply Integration By Parts

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

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

\int - 3 y^{2} (- 3 y + 3) d y = - 3 (y^{3} (- y + 1) + \frac{y ^{4}}{4})

= \frac{1}{2} y^{2} (3 - 3 y)^{2} - (- 3 (y^{3} (- y + 1) + \frac{y ^{4}}{4}))

Simplify

\frac{1}{2} y^{2} (3 - 3 y)^{2} - (- 3 (y^{3} (- y + 1) + \frac{y ^{4}}{4})) : \frac{3}{2} y^{4} + \frac{3 y ^{4}}{4} - 6 y^{3} + \frac{9}{2} y^{2}

= \frac{3}{2} y^{4} + \frac{3 y ^{4}}{4} - 6 y^{3} + \frac{9}{2} y^{2}

Add a constant to the solution

= \frac{3}{2} y^{4} + \frac{3 y ^{4}}{4} - 6 y^{3} + \frac{9}{2} y^{2} + C

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What is the integral of y(3-3y)^2 ?
The integral of y(3-3y)^2 is 3/2 y^4+(3y^4)/4-6y^3+9/2 y^2+C