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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

\int_{- 1}^{1} (y^{2} - 2) - e^{y} d y

\frac{e ^{x^{2}} - 1}{x}

n \to \infty lim (- n^{2})

\frac{d}{d x} (2 x^{3} (1 - x^{2})^{4})

x y^{^{'}} - 2 y + 8 = 0

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