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

The integral of 5x(2x+5)(x-3) is 5((x^4)/2-(x^3)/3-(15x^2)/2)+C

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

\int 5 x (2 x + 5) (x - 3) d x

5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2}) + C

Solution steps

\int 5 x (2 x + 5) (x - 3) d x

Take the constant out:

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

= 5 \cdot \int x (2 x + 5) (x - 3) d x

Expand

x (2 x + 5) (x - 3) : 2 x^{3} - x^{2} - 15 x

= 5 \cdot \int 2 x^{3} - x^{2} - 15 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

= 5 (\int 2 x^{3} d x - \int x^{2} d x - \int 15 x d x)

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

\int 15 x d x = \frac{15 x ^{2}}{2}

= 5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2})

Add a constant to the solution

= 5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2}) + C

t^{2} y^{^{'}} + t y = 7, y (1) = 4

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

x \to \infty lim (\frac{0}{1})

\frac{\partial}{\partial x} (\frac{( 9 y + 7 z )}{x + u})

\int \frac{3 r ^{2}}{r + 5} d r

What is the integral of 5x(2x+5)(x-3) ?
The integral of 5x(2x+5)(x-3) is 5((x^4)/2-(x^3)/3-(15x^2)/2)+C