What is the integral of 6x(3x^2-1) ?

The integral of 6x(3x^2-1) is 6((3x^4)/4-(x^2)/2)+C

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integral of 6x(3x^2-1)

\int 6 x (3 x^{2} - 1) d x

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

Solution steps

\int 6 x (3 x^{2} - 1) d x

Take the constant out:

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

= 6 \cdot \int x (3 x^{2} - 1) d x

Expand

x (3 x^{2} - 1) : 3 x^{3} - x

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

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

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

\int x d x = \frac{x ^{2}}{2}

= 6 (\frac{3 x ^{4}}{4} - \frac{x ^{2}}{2})

Add a constant to the solution

= 6 (\frac{3 x ^{4}}{4} - \frac{x ^{2}}{2}) + C

\frac{1}{x ^{2} + 9}

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What is the integral of 6x(3x^2-1) ?
The integral of 6x(3x^2-1) is 6((3x^4)/4-(x^2)/2)+C