What is the integral of x(x+2)(4x-1) ?

The integral of x(x+2)(4x-1) is x^4+(7x^3)/3-x^2+C

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integral of x(x+2)(4x-1)

\int x (x + 2) (4 x - 1) d x

x^{4} + \frac{7 x ^{3}}{3} - x^{2} + C

Solution steps

\int x (x + 2) (4 x - 1) d x

Expand

x (x + 2) (4 x - 1) : 4 x^{3} + 7 x^{2} - 2 x

= \int 4 x^{3} + 7 x^{2} - 2 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

= \int 4 x^{3} d x + \int 7 x^{2} d x - \int 2 x d x

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

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

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

= x^{4} + \frac{7 x ^{3}}{3} - x^{2}

Add a constant to the solution

= x^{4} + \frac{7 x ^{3}}{3} - x^{2} + C

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What is the integral of x(x+2)(4x-1) ?
The integral of x(x+2)(4x-1) is x^4+(7x^3)/3-x^2+C