solve ∫_0 ^( 1) ((1−x^2 )/((1+x^2 )(√(1+x^4 )))) dx ?


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Question Number 126266 by bramlexs22 last updated on 18/Dec/20

$s o l v e \int_{0}^{1} \frac{1 - x^{2}}{(1 + x^{2}) \sqrt{1 + x^{4}}} d x ?$
	Answered by liberty last updated on 19/Dec/20
	$N=∫_0 ^( 1) ((x^2 (−1+(1/x^2 )))/(x(x+(1/x))(√(x^2 ((1/x^2 )+x^2 ))))) dx N=∫_0 ^( 1) ((−1+(1/x^2 ))/((x+(1/x))(√((x+(1/x))^2 −((√2))^2 )))) dx [ put x+(1/x)=w→ { ((w=1)),((w=∞)) :}] N=∫_∞ ^( 1) ((−1)/(w(√(w^2 −((√2))^2 )))) dw N=∫_1 ^( ∞) (dw/(w(√(w^2 −((√2))^2 )))) = (1/( (√2))) [arcsec ((w/( (√2)))) ]_1 ^∞ N=(1/( (√2)))((π/2)−(π/4))= (π/(4(√2)))$
	$N = \int_{0}^{1} \frac{x^{2} (- 1 + \frac{1}{x^{2}})}{x (x + \frac{1}{x}) \sqrt{x^{2} (\frac{1}{x^{2}} + x^{2})}} d x$ $N = \int_{0}^{1} \frac{- 1 + \frac{1}{x^{2}}}{(x + \frac{1}{x}) \sqrt{{(x + \frac{1}{x})}^{2} - {(\sqrt{2})}^{2}}} d x$ $[p u t x + \frac{1}{x} = w \to {\begin{cases} w = 1 \\ w = \infty \end{cases}]$ $N = \int_{\infty}^{1} \frac{- 1}{w \sqrt{w^{2} - {(\sqrt{2})}^{2}}} d w$ $N = \int_{1}^{\infty} \frac{d w}{w \sqrt{w^{2} - {(\sqrt{2})}^{2}}} = \frac{1}{\sqrt{2}} {[arcsec (\frac{w}{\sqrt{2}})]}_{1}^{\infty}$ $N = \frac{1}{\sqrt{2}} (\frac{π}{2} - \frac{π}{4}) = \frac{π}{4 \sqrt{2}}$
	Commented by bramlexs22 last updated on 19/Dec/20

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