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

The integral of 2(2x^3-1)^3x^2 is 2((2x^{12})/3-(4x^9)/3+x^6-(x^3)/3)+C

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

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

2 (\frac{2 x ^{12}}{3} - \frac{4 x ^{9}}{3} + x^{6} - \frac{x ^{3}}{3}) + C

Solution steps

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

Take the constant out:

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

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

Expand

(2 x^{3} - 1)^{3} x^{2} : 8 x^{11} - 12 x^{8} + 6 x^{5} - x^{2}

= 2 \cdot \int 8 x^{11} - 12 x^{8} + 6 x^{5} - 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

= 2 (\int 8 x^{11} d x - \int 12 x^{8} d x + \int 6 x^{5} d x - \int x^{2} d x)

\int 8 x^{11} d x = \frac{2 x ^{12}}{3}

\int 12 x^{8} d x = \frac{4 x ^{9}}{3}

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

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

= 2 (\frac{2 x ^{12}}{3} - \frac{4 x ^{9}}{3} + x^{6} - \frac{x ^{3}}{3})

Add a constant to the solution

= 2 (\frac{2 x ^{12}}{3} - \frac{4 x ^{9}}{3} + x^{6} - \frac{x ^{3}}{3}) + C

What is the integral of 2(2x^3-1)^3x^2 ?
The integral of 2(2x^3-1)^3x^2 is 2((2x^{12})/3-(4x^9)/3+x^6-(x^3)/3)+C