What is the integral of 2x(x^2-3x+4) ?

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

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integral of 2x(x^2-3x+4)

\int 2 x (x^{2} - 3 x + 4) d x

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

Solution steps

\int 2 x (x^{2} - 3 x + 4) d x

Take the constant out:

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

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

Expand

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

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

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

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

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

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

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

Add a constant to the solution

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

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