What is the integral of 6x(x^3-1)^3 ?

The integral of 6x(x^3-1)^3 is 6((x^{11}}{11}-\frac{3x^8)/8+(3x^5)/5-(x^2)/2)+C

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integral of 6x(x^3-1)^3

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

6 (\frac{x ^{11}}{11} - \frac{3 x ^{8}}{8} + \frac{3 x ^{5}}{5} - \frac{x ^{2}}{2}) + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int x (x^{3} - 1)^{3} d x

Expand

x (x^{3} - 1)^{3} : x^{10} - 3 x^{7} + 3 x^{4} - x

= 6 \cdot \int x^{10} - 3 x^{7} + 3 x^{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

= 6 (\int x^{10} d x - \int 3 x^{7} d x + \int 3 x^{4} d x - \int x d x)

\int x^{10} d x = \frac{x ^{11}}{11}

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

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

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

= 6 (\frac{x ^{11}}{11} - \frac{3 x ^{8}}{8} + \frac{3 x ^{5}}{5} - \frac{x ^{2}}{2})

Add a constant to the solution

= 6 (\frac{x ^{11}}{11} - \frac{3 x ^{8}}{8} + \frac{3 x ^{5}}{5} - \frac{x ^{2}}{2}) + C

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What is the integral of 6x(x^3-1)^3 ?
The integral of 6x(x^3-1)^3 is 6((x^{11}}{11}-\frac{3x^8)/8+(3x^5)/5-(x^2)/2)+C