What is the integral of (2x^5-20x)^3(x^4-2) ?

The integral of (2x^5-20x)^3(x^4-2) is (2x^{20})/5-16x^{16}+240x^{12}-1600x^8+4000x^4+C

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integral of (2x^5-20x)^3(x^4-2)

\int (2 x^{5} - 20 x)^{3} (x^{4} - 2) d x

\frac{2 x ^{20}}{5} - 16 x^{16} + 240 x^{12} - 1600 x^{8} + 4000 x^{4} + C

Solution steps

\int (2 x^{5} - 20 x)^{3} (x^{4} - 2) d x

Expand

(2 x^{5} - 20 x)^{3} (x^{4} - 2) : 8 x^{19} - 256 x^{15} + 2880 x^{11} - 12800 x^{7} + 16000 x^{3}

= \int 8 x^{19} - 256 x^{15} + 2880 x^{11} - 12800 x^{7} + 16000 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 8 x^{19} d x - \int 256 x^{15} d x + \int 2880 x^{11} d x - \int 12800 x^{7} d x + \int 16000 x^{3} d x

\int 8 x^{19} d x = \frac{2 x ^{20}}{5}

\int 256 x^{15} d x = 16 x^{16}

\int 2880 x^{11} d x = 240 x^{12}

\int 12800 x^{7} d x = 1600 x^{8}

\int 16000 x^{3} d x = 4000 x^{4}

= \frac{2 x ^{20}}{5} - 16 x^{16} + 240 x^{12} - 1600 x^{8} + 4000 x^{4}

Add a constant to the solution

= \frac{2 x ^{20}}{5} - 16 x^{16} + 240 x^{12} - 1600 x^{8} + 4000 x^{4} + C

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What is the integral of (2x^5-20x)^3(x^4-2) ?
The integral of (2x^5-20x)^3(x^4-2) is (2x^{20})/5-16x^{16}+240x^{12}-1600x^8+4000x^4+C