What is the integral of 3y^2(y^3-4) ?

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

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integral of 3y^2(y^3-4)

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int y^{2} (y^{3} - 4) d y

Apply Integration By Parts

: \frac{1}{3} y^{3} (y^{3} - 4) - \int y^{5} d y

= 3 (\frac{1}{3} y^{3} (y^{3} - 4) - \int y^{5} d y)

\int y^{5} d y = \frac{y ^{6}}{6}

= 3 (\frac{1}{3} y^{3} (y^{3} - 4) - \frac{y ^{6}}{6})

Simplify

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

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

Add a constant to the solution

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

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