What is the integral of (2x^5-2x)^3(10x^4-2) ?

The integral of (2x^5-2x)^3(10x^4-2) is 4x^{20}-16x^{16}+24x^{12}-16x^8+4x^4+C

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integral of (2x^5-2x)^3(10x^4-2)

\int (2 x^{5} - 2 x)^{3} (10 x^{4} - 2) d x

4 x^{20} - 16 x^{16} + 24 x^{12} - 16 x^{8} + 4 x^{4} + C

Solution steps

\int (2 x^{5} - 2 x)^{3} (10 x^{4} - 2) d x

Expand

(2 x^{5} - 2 x)^{3} (10 x^{4} - 2) : 80 x^{19} - 256 x^{15} + 288 x^{11} - 128 x^{7} + 16 x^{3}

= \int 80 x^{19} - 256 x^{15} + 288 x^{11} - 128 x^{7} + 16 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 80 x^{19} d x - \int 256 x^{15} d x + \int 288 x^{11} d x - \int 128 x^{7} d x + \int 16 x^{3} d x

\int 80 x^{19} d x = 4 x^{20}

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

\int 288 x^{11} d x = 24 x^{12}

\int 128 x^{7} d x = 16 x^{8}

\int 16 x^{3} d x = 4 x^{4}

= 4 x^{20} - 16 x^{16} + 24 x^{12} - 16 x^{8} + 4 x^{4}

Add a constant to the solution

= 4 x^{20} - 16 x^{16} + 24 x^{12} - 16 x^{8} + 4 x^{4} + C

\frac{d}{d x} (\frac{x ^{3}}{e ^{x}})

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

f (x) = \frac{1}{r + 3 e ^{r}}

(8 + x) y^{^{'}} = 9 y

\frac{\partial}{\partial x} (y^{2} + 2 x y + 7 y + 7 x)

What is the integral of (2x^5-2x)^3(10x^4-2) ?
The integral of (2x^5-2x)^3(10x^4-2) is 4x^{20}-16x^{16}+24x^{12}-16x^8+4x^4+C