What is the integral of x^3(x^4-2)^3 ?

The integral of x^3(x^4-2)^3 is (x^{16}}{16}-\frac{x^{12})/2+(3x^8)/2-2x^4+C

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integral of x^3(x^4-2)^3

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

\frac{x ^{16}}{16} - \frac{x ^{12}}{2} + \frac{3 x ^{8}}{2} - 2 x^{4} + C

Solution steps

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

Expand

x^{3} (x^{4} - 2)^{3} : x^{15} - 6 x^{11} + 12 x^{7} - 8 x^{3}

= \int x^{15} - 6 x^{11} + 12 x^{7} - 8 x^{3} 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^{15} d x - \int 6 x^{11} d x + \int 12 x^{7} d x - \int 8 x^{3} d x

\int x^{15} d x = \frac{x ^{16}}{16}

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

\int 12 x^{7} d x = \frac{3 x ^{8}}{2}

\int 8 x^{3} d x = 2 x^{4}

= \frac{x ^{16}}{16} - \frac{x ^{12}}{2} + \frac{3 x ^{8}}{2} - 2 x^{4}

Add a constant to the solution

= \frac{x ^{16}}{16} - \frac{x ^{12}}{2} + \frac{3 x ^{8}}{2} - 2 x^{4} + C

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What is the integral of x^3(x^4-2)^3 ?
The integral of x^3(x^4-2)^3 is (x^{16}}{16}-\frac{x^{12})/2+(3x^8)/2-2x^4+C