∫_0 ^(π/2) ((1−x^4 )/(1+x^4 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 87121 by M±th+et£s last updated on 03/Apr/20

$\int_{0}^{\frac{π}{2}} \frac{1 - x^{4}}{1 + x^{4}} d x$
	Answered by redmiiuser last updated on 03/Apr/20

	${(1 + x^{4})}^{- 1}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n}$ $(1 - x^{4}) {(1 + x^{4})}^{- 1}$ $= (1 - x^{4}) . (\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n})$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n} - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n + 4}$ $\int_{0}^{\frac{π}{2}} \frac{1 - x^{4}}{1 + x^{4}} d x$ $Missing \left or extra \right$ $= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . {(\frac{π}{2})}^{4 n + 1}}{4 n + 1} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . {(\frac{π}{2})}^{4 n + 5}}{4 n + 5}$
	Commented by redmiiuser last updated on 03/Apr/20

	$p l s p l s p l s c h e c k$
	Commented by M±th+et£s last updated on 03/Apr/20

	$i t s r i g h t s i r t h a n k y o u$
	Commented by redmiiuser last updated on 03/Apr/20

	$w e l c o m e m i s t e r$
	Commented by MJS last updated on 03/Apr/20

	${what}^{'} s the value of your formula ?$
	Commented by redmiiuser last updated on 03/Apr/20

	$f r o m t h e c l o s e d f o r m s$ $i t s m u c h e f f i c e n t t o$ $i n s e r t t h e b o r d e r s$
	Commented by redmiiuser last updated on 03/Apr/20

	$A n y w a y T h a n k s S i r!$
	Commented by mr W last updated on 03/Apr/20

	$a s m u c h a s o n e c a n n o t g e t t h e v a l u e$ $o f t h e i n f i n i t e s e r i e s, i t h i n k s u c h a$ $s o l u t i o n i s n o t s o u s e f u l .$
	Commented by redmiiuser last updated on 03/Apr/20

	$T h e p r o c e s s i s n o t$ $a c l o s e d f o r m b u t y e t$ $i t i s e f f i c e n t .$ $A n y w a y T h a n k s f o r$ $y o u r c o m m e n t .$
	Answered by MJS last updated on 03/Apr/20

	$\int \frac{1 - x^{4}}{1 + x^{4}} d x = \int (\frac{2}{x^{4} + 1} - x) d x =$ $= - \int d x - \frac{\sqrt{2}}{2} \int \frac{x - \sqrt{2}}{x^{2} - \sqrt{2} x + 1} d x + \frac{\sqrt{2}}{2} \int \frac{x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} d x =$ $= - x -$ $- \frac{\sqrt{2}}{4} \ln (x^{2} - \sqrt{2} x + 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} x - 1) +$ $+ \frac{\sqrt{2}}{4} \ln (x^{2} + \sqrt{2} x + 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} x + 1) + C =$ $= - x + \frac{\sqrt{2}}{4} (\ln \frac{x^{2} + \sqrt{2} x + 1}{x^{2} - \sqrt{2} x + 1} + 2 (\arctan (\sqrt{2} x - 1) + \arctan (\sqrt{2} x + 1))) + C =$
	Commented by redmiiuser last updated on 03/Apr/20

	$s i r i t s a d e f i n i t e$ $i n t e g r a t i o n .$
	Commented by MJS last updated on 03/Apr/20

	$you can insert the borders$
	Commented by redmiiuser last updated on 03/Apr/20

	$s i r c a n y o u m a k e t h e$ $p r o c e s s m o r e s m o o t h .$
	Commented by M±th+et£s last updated on 03/Apr/20

	$t h a n k y o u i t s e a s y t o i n s e r t t h e b o r d e r s n o w$
	Commented by redmiiuser last updated on 03/Apr/20

	$G o d b l e s s y o u S i r .$ $H a v e a n i c e d a y .$
	Answered by TANMAY PANACEA. last updated on 03/Apr/20
	$(−1)∫_0 ^(π/2) ((x^4 +1−2)/(1+x^4 ))dx ∫_0 ^(π/2) (2/(1+x^4 ))−∫_0 ^(π/2) dx ∫_0 ^(π/2) ((1+(1/x^2 )−(1−(1/x^2 )))/(x^2 +(1/x^2 )))−∫_0 ^(π/2) dx ∫_0 ^(π/2) ((d(x−(1/x)))/((x−(1/x))^2 +2))−∫_0 ^(π/2) ((d(x+(1/x)))/((x+(1/x))^2 −2))−∫_0 ^(π/2) dx ∣(1/(√2))tan^(−1) (((x−(1/x))/(√2)))−(1/(2(√2)))ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))−x∣_0 ^(π/2) {(1/(√2))tan^(−1) ((((π/2)−(2/π))/(√2)))−(1/(2(√2)))ln((((π/2)+(2/π)−(√2))/((π/2)+(2/π)+(√2))))−(π/2)}− {(1/(√2))tan^(−1) (−∞)−(1/(2(√2)))ln(((0+1−0)/(0+1−0)))−0} =(1/(√2))tan^(−1) (((π^2 −4)/(2(√2) π)))−(1/(2(√2)))ln(((π^2 +2^2 −2(√2) π)/(π^2 +2^2 +2(√2)π)))−(π/2)−(1/(√2))(((−π)/2)) =(1/(√2))tan^(−1) (((π^2 −4)/(2(√(2π)))))−(1/(2(√2)))ln(((π^2 +4−2(√2) π)/(π^2 +4+2(√2) π)))−(π/2)(1+(1/(√2))) ★ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))→ln(((x^2 +1−(√2) x)/(x^2 +1+(√2) x))) ★$
	$(- 1) \int_{0}^{\frac{π}{2}} \frac{x^{4} + 1 - 2}{1 + x^{4}} d x$ $\int_{0}^{\frac{π}{2}} \frac{2}{1 + x^{4}} - \int_{0}^{\frac{π}{2}} d x$ $\int_{0}^{\frac{π}{2}} \frac{1 + \frac{1}{x^{2}} - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} - \int_{0}^{\frac{π}{2}} d x$ $\int_{0}^{\frac{π}{2}} \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 2} - \int_{0}^{\frac{π}{2}} \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 2} - \int_{0}^{\frac{π}{2}} d x$ $∣ \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) - x ∣_{0}^{\frac{π}{2}}$ ${\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{\frac{π}{2} - \frac{2}{π}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{\frac{π}{2} + \frac{2}{π} - \sqrt{2}}{\frac{π}{2} + \frac{2}{π} + \sqrt{2}}) - \frac{π}{2}} -$ ${\frac{1}{\sqrt{2}} {t a n}^{- 1} (- \infty) - \frac{1}{2 \sqrt{2}} l n (\frac{0 + 1 - 0}{0 + 1 - 0}) - 0}$ $= \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{π^{2} - 4}{2 \sqrt{2} π}) - \frac{1}{2 \sqrt{2}} l n (\frac{π^{2} + 2^{2} - 2 \sqrt{2} π}{π^{2} + 2^{2} + 2 \sqrt{2} π}) - \frac{π}{2} - \frac{1}{\sqrt{2}} (\frac{- π}{2})$ $= \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{π^{2} - 4}{2 \sqrt{2 π}}) - \frac{1}{2 \sqrt{2}} l n (\frac{π^{2} + 4 - 2 \sqrt{2} π}{π^{2} + 4 + 2 \sqrt{2} π}) - \frac{π}{2} (1 + \frac{1}{\sqrt{2}})$ $★ l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) \to l n (\frac{x^{2} + 1 - \sqrt{2} x}{x^{2} + 1 + \sqrt{2} x}) ★$
	Commented by TANMAY PANACEA. last updated on 03/Apr/20

	$m o s t w e l c o m e s i r . . .$
	Commented by M±th+et£s last updated on 03/Apr/20

	$t h a n k y o u s i r$
	Commented by peter frank last updated on 03/Apr/20

	$t h a n k y o u b o t h$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum