What is the integral of 5x(x^2-3x) ?

The integral of 5x(x^2-3x) is 5((x^4)/4-x^3)+C

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integral of 5x(x^2-3x)

\int 5 x (x^{2} - 3 x) d x

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

Solution steps

\int 5 x (x^{2} - 3 x) d x

Take the constant out:

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

= 5 \cdot \int x (x^{2} - 3 x) d x

Expand

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

= 5 \cdot \int x^{3} - 3 x^{2} 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

= 5 (\int x^{3} d x - \int 3 x^{2} d x)

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

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

= 5 (\frac{x ^{4}}{4} - x^{3})

Add a constant to the solution

= 5 (\frac{x ^{4}}{4} - x^{3}) + C

What is the integral of 5x(x^2-3x) ?
The integral of 5x(x^2-3x) is 5((x^4)/4-x^3)+C