What is the integral of 3x(6-x^2) ?

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

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integral of 3x(6-x^2)

\int 3 x (6 - x^{2}) d x

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

Solution steps

\int 3 x (6 - x^{2}) d x

Take the constant out:

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

= 3 \cdot \int x (6 - x^{2}) d x

Expand

x (6 - x^{2}) : 6 x - x^{3}

= 3 \cdot \int 6 x - x^{3} d x

Apply the Sum Rule:

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

= 3 (\int 6 x d x - \int x^{3} d x)

\int 6 x d x = 3 x^{2}

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

= 3 (3 x^{2} - \frac{x ^{4}}{4})

Add a constant to the solution

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

\int (- 1)^{n} x^{n} d x

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

\frac{\partial}{\partial x} (2 y + y e^{x})

\frac{d x}{d t} = - 3 x

h \to 0 lim (\frac{e ^{2 h} - 1}{h})

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