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

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

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

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

3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3}) + C

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (x - 2) (x + 3) : x^{4} + x^{3} - 6 x^{2}

= 3 \cdot \int x^{4} + x^{3} - 6 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^{4} d x + \int x^{3} d x - \int 6 x^{2} d x)

\int x^{4} d x = \frac{x ^{5}}{5}

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

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

= 3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3})

Add a constant to the solution

= 3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3}) + C

(1 + x^{2}) \frac{d y}{d x} - (2 x y + x^{2} + x^{4}) = 0

\frac{d x}{d t} - \frac{2}{3 t} x = 4 t x^{- 5}

\frac{d}{d x} (x^{2} (e^{x} - 1))

x \to \frac{π}{2} lim (2 sec (x))

\int_{1}^{2} 3 \frac{ln ( x )}{x ^{2}} d x

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