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

The integral of (1-x^3)^2 is x-(x^4)/2+(x^7)/7+C

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integral of (1-x^3)^2

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

x - \frac{x ^{4}}{2} + \frac{x ^{7}}{7} + C

Solution steps

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

Expand

(1 - x^{3})^{2} : 1 - 2 x^{3} + x^{6}

= \int 1 - 2 x^{3} + x^{6} 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 1 d x - \int 2 x^{3} d x + \int x^{6} d x

\int 1 d x = x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int x^{6} d x = \frac{x ^{7}}{7}

= x - \frac{x ^{4}}{2} + \frac{x ^{7}}{7}

Add a constant to the solution

= x - \frac{x ^{4}}{2} + \frac{x ^{7}}{7} + C

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What is the integral of (1-x^3)^2 ?
The integral of (1-x^3)^2 is x-(x^4)/2+(x^7)/7+C