What is the integral of x^2(2x-1)(x-6) ?

The integral of x^2(2x-1)(x-6) is (2x^5)/5-(13x^4)/4+2x^3+C

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integral of x^2(2x-1)(x-6)

\int x^{2} (2 x - 1) (x - 6) d x

\frac{2 x ^{5}}{5} - \frac{13 x ^{4}}{4} + 2 x^{3} + C

Solution steps

\int x^{2} (2 x - 1) (x - 6) d x

Expand

x^{2} (2 x - 1) (x - 6) : 2 x^{4} - 13 x^{3} + 6 x^{2}

= \int 2 x^{4} - 13 x^{3} + 6 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 2 x^{4} d x - \int 13 x^{3} d x + \int 6 x^{2} d x

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int 13 x^{3} d x = \frac{13 x ^{4}}{4}

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

= \frac{2 x ^{5}}{5} - \frac{13 x ^{4}}{4} + 2 x^{3}

Add a constant to the solution

= \frac{2 x ^{5}}{5} - \frac{13 x ^{4}}{4} + 2 x^{3} + C

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What is the integral of x^2(2x-1)(x-6) ?
The integral of x^2(2x-1)(x-6) is (2x^5)/5-(13x^4)/4+2x^3+C