What is the integral of 20(x-3)(x+2)^3 ?

The integral of 20(x-3)(x+2)^3 is 20((x^5)/5+(3x^4)/4-2x^3-14x^2-24x)+C

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integral of 20(x-3)(x+2)^3

integral

20 (x - 3) (x + 2)^{3}

20 (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 2 x^{3} - 14 x^{2} - 24 x) + C

Solution steps

\int 20 (x - 3) (x + 2)^{3} d x

Take the constant out:

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

= 20 \cdot \int (x - 3) (x + 2)^{3} d x

Expand

(x - 3) (x + 2)^{3} : x^{4} + 3 x^{3} - 6 x^{2} - 28 x - 24

= 20 \cdot \int x^{4} + 3 x^{3} - 6 x^{2} - 28 x - 24 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

= 20 (\int x^{4} d x + \int 3 x^{3} d x - \int 6 x^{2} d x - \int 28 x d x - \int 24 d x)

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

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

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

\int 28 x d x = 14 x^{2}

\int 24 d x = 24 x

= 20 (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 2 x^{3} - 14 x^{2} - 24 x)

Add a constant to the solution

= 20 (\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 2 x^{3} - 14 x^{2} - 24 x) + C

in t e g r a l \frac{1}{x ^{\frac{1}{2}}}

in t e g r a l y = x^{2}

in t e g r a l \frac{x}{( 1 + x ^{3} )}

in t e g r a l \frac{e ^{x}}{( e ^{x} + 1 ) ^{2}}

in t e g r a l x^{2} a^{2} + x^{2}

What is the integral of 20(x-3)(x+2)^3 ?
The integral of 20(x-3)(x+2)^3 is 20((x^5)/5+(3x^4)/4-2x^3-14x^2-24x)+C