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

The integral of (x^2-4)(x^3-4x)^3 is (x^{12}}{12}-\frac{8x^{10})/5+12x^8-(128x^6)/3+64x^4+C

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

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

\frac{x ^{12}}{12} - \frac{8 x ^{10}}{5} + 12 x^{8} - \frac{128 x ^{6}}{3} + 64 x^{4} + C

Solution steps

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

Expand

(x^{2} - 4) (x^{3} - 4 x)^{3} : x^{11} - 16 x^{9} + 96 x^{7} - 256 x^{5} + 256 x^{3}

= \int x^{11} - 16 x^{9} + 96 x^{7} - 256 x^{5} + 256 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^{11} d x - \int 16 x^{9} d x + \int 96 x^{7} d x - \int 256 x^{5} d x + \int 256 x^{3} d x

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

\int 16 x^{9} d x = \frac{8 x ^{10}}{5}

\int 96 x^{7} d x = 12 x^{8}

\int 256 x^{5} d x = \frac{128 x ^{6}}{3}

\int 256 x^{3} d x = 64 x^{4}

= \frac{x ^{12}}{12} - \frac{8 x ^{10}}{5} + 12 x^{8} - \frac{128 x ^{6}}{3} + 64 x^{4}

Add a constant to the solution

= \frac{x ^{12}}{12} - \frac{8 x ^{10}}{5} + 12 x^{8} - \frac{128 x ^{6}}{3} + 64 x^{4} + C

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What is the integral of (x^2-4)(x^3-4x)^3 ?
The integral of (x^2-4)(x^3-4x)^3 is (x^{12}}{12}-\frac{8x^{10})/5+12x^8-(128x^6)/3+64x^4+C