What is the integral of (x^8-4x^2)^3(x^7-x) ?

The integral of (x^8-4x^2)^3(x^7-x) is (x^{32}}{32}-\frac{x^{26})/2+3x^{20}-8x^{14}+8x^8+C

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integral of (x^8-4x^2)^3(x^7-x)

\int (x^{8} - 4 x^{2})^{3} (x^{7} - x) d x

\frac{x ^{32}}{32} - \frac{x ^{26}}{2} + 3 x^{20} - 8 x^{14} + 8 x^{8} + C

Solution steps

\int (x^{8} - 4 x^{2})^{3} (x^{7} - x) d x

Expand

(x^{8} - 4 x^{2})^{3} (x^{7} - x) : x^{31} - 13 x^{25} + 60 x^{19} - 112 x^{13} + 64 x^{7}

= \int x^{31} - 13 x^{25} + 60 x^{19} - 112 x^{13} + 64 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 x^{31} d x - \int 13 x^{25} d x + \int 60 x^{19} d x - \int 112 x^{13} d x + \int 64 x^{7} d x

\int x^{31} d x = \frac{x ^{32}}{32}

\int 13 x^{25} d x = \frac{x ^{26}}{2}

\int 60 x^{19} d x = 3 x^{20}

\int 112 x^{13} d x = 8 x^{14}

\int 64 x^{7} d x = 8 x^{8}

= \frac{x ^{32}}{32} - \frac{x ^{26}}{2} + 3 x^{20} - 8 x^{14} + 8 x^{8}

Add a constant to the solution

= \frac{x ^{32}}{32} - \frac{x ^{26}}{2} + 3 x^{20} - 8 x^{14} + 8 x^{8} + C

\frac{\partial}{\partial z} (x e^{y})

y = \frac{1}{2 x + 7}

f (x) = 2 x + \frac{5}{x}

ln (2 - x^{2})

(\frac{x + 1}{x - 1})^{2}

What is the integral of (x^8-4x^2)^3(x^7-x) ?
The integral of (x^8-4x^2)^3(x^7-x) is (x^{32}}{32}-\frac{x^{26})/2+3x^{20}-8x^{14}+8x^8+C