What is the integral of 6x^2(3-2x^3)^3 ?

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

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integral of 6x^2(3-2x^3)^3

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

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

Solution steps

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

Expand

x^{2} (3 - 2 x^{3})^{3} : 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11}

= 6 \cdot \int 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11} 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 27 x^{2} d x - \int 54 x^{5} d x + \int 36 x^{8} d x - \int 8 x^{11} d x)

\int 27 x^{2} d x = 9 x^{3}

\int 54 x^{5} d x = 9 x^{6}

\int 36 x^{8} d x = 4 x^{9}

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

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

Add a constant to the solution

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

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 58} d x

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\int (\frac{x ^{3}}{3} - \frac{x ^{4}}{2} + \frac{x ^{5}}{5}) d x

\int \frac{1}{x + 3 x - 4} d x

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