Given-f-x-1-x-x-4-1-x-4-2-then-1-2-1-x-2-f-x-dx-

Question Number 128221 by john_santu last updated on 05/Jan/21

G i v e n f (x + \frac{1}{x}) = x^{4} - \frac{1}{x^{4}} + 2

t h e n \int_{1}^{2} (1 - x^{- 2}) f (x) d x =

Answered by liberty last updated on 05/Jan/21

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

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

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

t h e n f (x + \frac{1}{x}) = [{(x + \frac{1}{x})}^{2} - 2] (x + \frac{1}{x}) \sqrt{{(x + \frac{1}{x})}^{2} - 4} + 2

o r f (x) = (x^{2} - 2) x \sqrt{x^{2} - 4} + 2

t h e n \int_{1}^{2} (\frac{x^{2} - 1}{x^{2}}) [x (x^{2} - 2) \sqrt{x^{2} - 4} + 2] d x =

\int_{1}^{2} \frac{(x^{2} - 1) (x^{2} - 2) \sqrt{x^{2} - 4}}{x} d x - \int_{1}^{2} \frac{2 (x^{2} - 1)}{x^{2}} d x

I_{1} = \int_{1}^{2} \frac{(x^{4} - 3 x^{2} + 2) \sqrt{x^{2} - 4}}{x} d x

I_{2} = \int_{1}^{2} 2 (1 - x^{- 2}) d x

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