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

The integral of (x^4-8x)^2(4x^3-8) is (x^{12})/3-8x^9+64x^6-(512x^3)/3+C

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

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

\frac{x ^{12}}{3} - 8 x^{9} + 64 x^{6} - \frac{512 x ^{3}}{3} + C

Solution steps

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

Expand

(x^{4} - 8 x)^{2} (4 x^{3} - 8) : 4 x^{11} - 72 x^{8} + 384 x^{5} - 512 x^{2}

= \int 4 x^{11} - 72 x^{8} + 384 x^{5} - 512 x^{2} 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 4 x^{11} d x - \int 72 x^{8} d x + \int 384 x^{5} d x - \int 512 x^{2} d x

\int 4 x^{11} d x = \frac{x ^{12}}{3}

\int 72 x^{8} d x = 8 x^{9}

\int 384 x^{5} d x = 64 x^{6}

\int 512 x^{2} d x = \frac{512 x ^{3}}{3}

= \frac{x ^{12}}{3} - 8 x^{9} + 64 x^{6} - \frac{512 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{12}}{3} - 8 x^{9} + 64 x^{6} - \frac{512 x ^{3}}{3} + C

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What is the integral of (x^4-8x)^2(4x^3-8) ?
The integral of (x^4-8x)^2(4x^3-8) is (x^{12})/3-8x^9+64x^6-(512x^3)/3+C