What is the integral of 3x^5(x^3-1)^4 ?

The integral of 3x^5(x^3-1)^4 is 3((x^{18}}{18}-(4x^{15})/(15)+\frac{x^{12})/2-(4x^9)/9+(x^6)/6)+C

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integral of 3x^5(x^3-1)^4

\int 3 x^{5} (x^{3} - 1)^{4} d x

3 (\frac{x ^{18}}{18} - \frac{4 x ^{15}}{15} + \frac{x ^{12}}{2} - \frac{4 x ^{9}}{9} + \frac{x ^{6}}{6}) + C

Solution steps

\int 3 x^{5} (x^{3} - 1)^{4} d x

Take the constant out:

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

= 3 \cdot \int x^{5} (x^{3} - 1)^{4} d x

Expand

x^{5} (x^{3} - 1)^{4} : x^{17} - 4 x^{14} + 6 x^{11} - 4 x^{8} + x^{5}

= 3 \cdot \int x^{17} - 4 x^{14} + 6 x^{11} - 4 x^{8} + x^{5} 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

= 3 (\int x^{17} d x - \int 4 x^{14} d x + \int 6 x^{11} d x - \int 4 x^{8} d x + \int x^{5} d x)

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

\int 4 x^{14} d x = \frac{4 x ^{15}}{15}

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

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

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

= 3 (\frac{x ^{18}}{18} - \frac{4 x ^{15}}{15} + \frac{x ^{12}}{2} - \frac{4 x ^{9}}{9} + \frac{x ^{6}}{6})

Add a constant to the solution

= 3 (\frac{x ^{18}}{18} - \frac{4 x ^{15}}{15} + \frac{x ^{12}}{2} - \frac{4 x ^{9}}{9} + \frac{x ^{6}}{6}) + C

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What is the integral of 3x^5(x^3-1)^4 ?
The integral of 3x^5(x^3-1)^4 is 3((x^{18}}{18}-(4x^{15})/(15)+\frac{x^{12})/2-(4x^9)/9+(x^6)/6)+C