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

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

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

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

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

Solution steps

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

Expand

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

= \int x^{2} - 3 x^{5} + 3 x^{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

= \int x^{2} d x - \int 3 x^{5} d x + \int 3 x^{8} d x - \int x^{11} d x

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

\int x^{11} d x = \frac{x ^{12}}{12}

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

Add a constant to the solution

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

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