What is the integral of (8x^3-4x)(x^4-x^2)^3 ?

The integral of (8x^3-4x)(x^4-x^2)^3 is (x^{16})/2-2x^{14}+3x^{12}-2x^{10}+(x^8)/2+C

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integral of (8x^3-4x)(x^4-x^2)^3

\int (8 x^{3} - 4 x) (x^{4} - x^{2})^{3} d x

\frac{x ^{16}}{2} - 2 x^{14} + 3 x^{12} - 2 x^{10} + \frac{x ^{8}}{2} + C

Solution steps

\int (8 x^{3} - 4 x) (x^{4} - x^{2})^{3} d x

Expand

(8 x^{3} - 4 x) (x^{4} - x^{2})^{3} : 8 x^{15} - 28 x^{13} + 36 x^{11} - 20 x^{9} + 4 x^{7}

= \int 8 x^{15} - 28 x^{13} + 36 x^{11} - 20 x^{9} + 4 x^{7} 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 8 x^{15} d x - \int 28 x^{13} d x + \int 36 x^{11} d x - \int 20 x^{9} d x + \int 4 x^{7} d x

\int 8 x^{15} d x = \frac{x ^{16}}{2}

\int 28 x^{13} d x = 2 x^{14}

\int 36 x^{11} d x = 3 x^{12}

\int 20 x^{9} d x = 2 x^{10}

\int 4 x^{7} d x = \frac{x ^{8}}{2}

= \frac{x ^{16}}{2} - 2 x^{14} + 3 x^{12} - 2 x^{10} + \frac{x ^{8}}{2}

Add a constant to the solution

= \frac{x ^{16}}{2} - 2 x^{14} + 3 x^{12} - 2 x^{10} + \frac{x ^{8}}{2} + C

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What is the integral of (8x^3-4x)(x^4-x^2)^3 ?
The integral of (8x^3-4x)(x^4-x^2)^3 is (x^{16})/2-2x^{14}+3x^{12}-2x^{10}+(x^8)/2+C