What is the integral of (2x^3-x+0)^3*(6x^2-1) ?

The integral of (2x^3-x+0)^3*(6x^2-1) is 4x^{12}-8x^{10}+6x^8-2x^6+(x^4)/4+C

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integral of (2x^3-x+0)^3*(6x^2-1)

\int (2 x^{3} - x + 0)^{3} \cdot (6 x^{2} - 1) d x

4 x^{12} - 8 x^{10} + 6 x^{8} - 2 x^{6} + \frac{x ^{4}}{4} + C

Solution steps

\int (2 x^{3} - x + 0)^{3} (6 x^{2} - 1) d x

Expand

(2 x^{3} - x + 0)^{3} (6 x^{2} - 1) : 48 x^{11} - 80 x^{9} + 48 x^{7} - 12 x^{5} + x^{3}

= \int 48 x^{11} - 80 x^{9} + 48 x^{7} - 12 x^{5} + x^{3} 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 48 x^{11} d x - \int 80 x^{9} d x + \int 48 x^{7} d x - \int 12 x^{5} d x + \int x^{3} d x

\int 48 x^{11} d x = 4 x^{12}

\int 80 x^{9} d x = 8 x^{10}

\int 48 x^{7} d x = 6 x^{8}

\int 12 x^{5} d x = 2 x^{6}

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

= 4 x^{12} - 8 x^{10} + 6 x^{8} - 2 x^{6} + \frac{x ^{4}}{4}

Add a constant to the solution

= 4 x^{12} - 8 x^{10} + 6 x^{8} - 2 x^{6} + \frac{x ^{4}}{4} + C

What is the integral of (2x^3-x+0)^3*(6x^2-1) ?
The integral of (2x^3-x+0)^3*(6x^2-1) is 4x^{12}-8x^{10}+6x^8-2x^6+(x^4)/4+C