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

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

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

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

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

Solution steps

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

Expand

(3 x^{3} - x) (2 - x) : 6 x^{3} - 3 x^{4} - 2 x + x^{2}

= \int 6 x^{3} - 3 x^{4} - 2 x + 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 6 x^{3} d x - \int 3 x^{4} d x - \int 2 x d x + \int x^{2} d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

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

= \frac{3 x ^{4}}{2} - \frac{3 x ^{5}}{5} - x^{2} + \frac{x ^{3}}{3}

Add a constant to the solution

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

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