What is the integral of (3x-1)^2(-2x) ?

The integral of (3x-1)^2(-2x) is -(9x^4)/2+4x^3-x^2+C

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integral of (3x-1)^2(-2x)

\int (3 x - 1)^{2} (- 2 x) d x

- \frac{9 x ^{4}}{2} + 4 x^{3} - x^{2} + C

Solution steps

\int (3 x - 1)^{2} (- 2 x) d x

Simplify

= \int - 2 x (3 x - 1)^{2} d x

Take the constant out:

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

= - 2 \cdot \int x (3 x - 1)^{2} d x

Expand

x (3 x - 1)^{2} : 9 x^{3} - 6 x^{2} + x

= - 2 \cdot \int 9 x^{3} - 6 x^{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

= - 2 (\int 9 x^{3} d x - \int 6 x^{2} d x + \int x d x)

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

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

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

= - 2 (\frac{9 x ^{4}}{4} - 2 x^{3} + \frac{x ^{2}}{2})

Simplify

- 2 (\frac{9 x ^{4}}{4} - 2 x^{3} + \frac{x ^{2}}{2}) : - \frac{9 x ^{4}}{2} + 4 x^{3} - x^{2}

= - \frac{9 x ^{4}}{2} + 4 x^{3} - x^{2}

Add a constant to the solution

= - \frac{9 x ^{4}}{2} + 4 x^{3} - x^{2} + C

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What is the integral of (3x-1)^2(-2x) ?
The integral of (3x-1)^2(-2x) is -(9x^4)/2+4x^3-x^2+C