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

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

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integral of 2x(x^2-3)

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

= 2 \cdot \int x^{3} - 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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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