What is the integral of 6x^5(1-x^6)^2 ?

The integral of 6x^5(1-x^6)^2 is x^6-x^{12}+(x^{18})/3+C

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integral of 6x^5(1-x^6)^2

\int 6 x^{5} (1 - x^{6})^{2} d x

x^{6} - x^{12} + \frac{x ^{18}}{3} + C

Solution steps

\int 6 x^{5} (1 - x^{6})^{2} d x

Take the constant out:

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

= 6 \cdot \int x^{5} (1 - x^{6})^{2} d x

Expand

x^{5} (1 - x^{6})^{2} : x^{5} - 2 x^{11} + x^{17}

= 6 \cdot \int x^{5} - 2 x^{11} + x^{17} 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 x^{5} d x - \int 2 x^{11} d x + \int x^{17} d x)

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

\int 2 x^{11} d x = \frac{x ^{12}}{6}

\int x^{17} d x = \frac{x ^{18}}{18}

= 6 (\frac{x ^{6}}{6} - \frac{x ^{12}}{6} + \frac{x ^{18}}{18})

Simplify

6 (\frac{x ^{6}}{6} - \frac{x ^{12}}{6} + \frac{x ^{18}}{18}) : x^{6} - x^{12} + \frac{x ^{18}}{3}

= x^{6} - x^{12} + \frac{x ^{18}}{3}

Add a constant to the solution

= x^{6} - x^{12} + \frac{x ^{18}}{3} + C

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What is the integral of 6x^5(1-x^6)^2 ?
The integral of 6x^5(1-x^6)^2 is x^6-x^{12}+(x^{18})/3+C