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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{9}{x + 1}

\frac{d y}{d t} = - 3 y + 9

\int y^{2} e^{x y^{2}} + 1 d x

\int_{0}^{\frac{π}{2}} cos (4 x) d x

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

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