What is the integral of (9-z^2)(6-2z) ?

The integral of (9-z^2)(6-2z) is 54z-9z^2-2z^3+(z^4)/2+C

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integral of (9-z^2)(6-2z)

\int (9 - z^{2}) (6 - 2 z) d z

54 z - 9 z^{2} - 2 z^{3} + \frac{z ^{4}}{2} + C

Solution steps

\int (9 - z^{2}) (6 - 2 z) d z

(9 - z^{2}) (6 - 2 z) = 2 (9 - z^{2}) (3 - z)

= \int 2 (9 - z^{2}) (3 - z) d z

Take the constant out:

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

= 2 \cdot \int (9 - z^{2}) (3 - z) d z

Expand

(9 - z^{2}) (3 - z) : 27 - 9 z - 3 z^{2} + z^{3}

= 2 \cdot \int 27 - 9 z - 3 z^{2} + z^{3} d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int 27 d z - \int 9 z d z - \int 3 z^{2} d z + \int z^{3} d z)

\int 27 d z = 27 z

\int 9 z d z = \frac{9 z ^{2}}{2}

\int 3 z^{2} d z = z^{3}

\int z^{3} d z = \frac{z ^{4}}{4}

= 2 (27 z - \frac{9 z ^{2}}{2} - z^{3} + \frac{z ^{4}}{4})

Simplify

2 (27 z - \frac{9 z ^{2}}{2} - z^{3} + \frac{z ^{4}}{4}) : 54 z - 9 z^{2} - 2 z^{3} + \frac{z ^{4}}{2}

= 54 z - 9 z^{2} - 2 z^{3} + \frac{z ^{4}}{2}

Add a constant to the solution

= 54 z - 9 z^{2} - 2 z^{3} + \frac{z ^{4}}{2} + C

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What is the integral of (9-z^2)(6-2z) ?
The integral of (9-z^2)(6-2z) is 54z-9z^2-2z^3+(z^4)/2+C