What is the integral of x(20x^2+60x) ?

The integral of x(20x^2+60x) is 5x^4+20x^3+C

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integral of x(20x^2+60x)

\int x (20 x^{2} + 60 x) d x

5 x^{4} + 20 x^{3} + C

Solution steps

\int x (20 x^{2} + 60 x) d x

Expand

x (20 x^{2} + 60 x) : 20 x^{3} + 60 x^{2}

= \int 20 x^{3} + 60 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

= \int 20 x^{3} d x + \int 60 x^{2} d x

\int 20 x^{3} d x = 5 x^{4}

\int 60 x^{2} d x = 20 x^{3}

= 5 x^{4} + 20 x^{3}

Add a constant to the solution

= 5 x^{4} + 20 x^{3} + C

t an g e n t x^{4} + 2 e^{x}

d er i v a t i v e \frac{x}{1 - \frac{1}{x ^{2}}}

a re a \frac{7}{x}, x = 1, x = 41

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

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

What is the integral of x(20x^2+60x) ?
The integral of x(20x^2+60x) is 5x^4+20x^3+C