x-6-64-4-2x-1-x-2-x-2-4-4x-1-x-2-4x-2-2x-1-1-2x-dx-

Question Number 171926 by Mikenice last updated on 21/Jun/22

\int (\frac{x^{- 6} - 64}{4 + 2 x^{- 1} + x^{- 2}} . \frac{x^{2}}{4 - 4 x^{- 1} + x^{- 2}} - \frac{4 x^{2} (2 x + 1)}{1 - 2 x}) d x

Commented by infinityaction last updated on 22/Jun/22

x^{2} + x + c

Commented by infinityaction last updated on 22/Jun/22

p = \frac{x^{- 6} - 64}{4 + 2 x^{- 1} + x^{- 2}} . \frac{x^{2}}{4 - 4 x^{- 1} + x^{- 2}} - \frac{4 x^{2} (2 x + 1)}{1 - 2 x}

p = \frac{\frac{1}{x^{6}} - 64}{4 + \frac{2}{x} + \frac{1}{x^{2}}} \cdot \frac{x^{2}}{4 - \frac{4}{x} + \frac{1}{x^{2}}} - \frac{4 x^{2}}{1 - 2 x}

p = \frac{(1 - 64 x^{4}) x^{2}}{x^{6} (4 x^{2} + 2 x + 1)} \cdot \frac{x^{4}}{(4 x^{2} - 4 x + 1)} - \frac{4 x^{2} (1 + 2 x)}{1 - 2 x}

1 - 64 x^{6} = 1^{3} - {(4 x^{2})}^{3}

1 - 64 x^{6} = (1 - 4 x^{2}) (1 + 16 x^{4} + 4 x^{2})

p = \frac{(1 - 4 x^{2}) (1 + 16 x^{4} + 4 x^{2})}{(4 x^{2} + 2 x + 1} \cdot \frac{1}{4 x^{2} - 4 x + 1} - \frac{4 x^{2} (2 x + 1)}{1 - 2 x}

p = \frac{- (2 x - 1) (1 + 2 x) (1 + 16 x^{4} 4 x^{2})}{4 x^{2} + 2 x + 1) {(2 x - 1)}^{2}} \cdot + \frac{4 x^{2} (1 + 2 x)}{2 x - 1}

p = \frac{2 x + 1}{2 x - 1} {- \frac{1 + 16 x^{4} + 4 x 2}{1 + 2 x + 4 x^{2}} + 4 x^{2}}

p = \frac{2 x + 1}{2 x - 1} {\frac{- 1 - 16 x^{4} - 4 x^{2} + 8 x^{3} + 4 x^{2} + 16 x^{4}}{1 + 2 x + 4 x^{2}}}

p = \frac{2 x + 1}{2 x - 1} {\frac{{(2 x)}^{3} - 1}{1 + 2 x + 4 x^{2}}}

p = \frac{2 x + 1}{2 x - 1} {\frac{(2 x - 1) (1 + 2 x + 4 x^{2})}{1 + 2 x + 4 x^{2}}}

p = 2 x + 1

\int p d x = \int (2 x + 1) d x

\int p d x = x^{2} + x + C

Commented by Mikenice last updated on 23/Jun/22

t h a n k s s i r