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Question Number 117087 by mnjuly1970 last updated on 09/Oct/20

$. . . n i c e c a l c u l u s . . .$ $p l e a s e e v a l u a t e ::$ $Ω = \int_{0}^{1} \frac{x^{4} + 1}{x^{6} + 1} d x = ? ? ?$ $m . n . 1970$
	Answered by TANMAY PANACEA last updated on 09/Oct/20

	$\int \frac{{(x^{2} + 1)}^{2} - 2 x^{2}}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x$ $\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x - \int \frac{2 x^{2}}{{(x^{3})}^{2} + 1} d x$ $\int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 1} d x - \frac{2}{3} \int \frac{d (x^{3})}{{(x^{3})}^{2} + 1}$ $\int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 1} - \frac{2}{3} \int \frac{d (x^{3})}{{(x^{3})}^{2} + 1}$ ${t a n}^{- 1} (x - \frac{1}{x}) - \frac{2}{3} {t a n}^{- 1} (x^{3}) + C$ $n o w p u t l i m i t$ $∣ {t a n}^{- 1} (x - \frac{1}{x}) - \frac{2}{3} {t a n}^{- 1} (x^{3}) ∣_{0}^{1}$ ${t a n}^{- 1} (0) - \frac{2}{3} {t a n}^{- 1} (1) - {t a n}^{- 1} (0 - \infty) + \frac{2}{3} {t a n}^{- 1} (0)$ $= \frac{π}{4} \times \frac{- 2}{3} + \frac{π}{2}$ $= \frac{π}{2} (1 - \frac{1}{3}) = \frac{π}{3}$
	Commented by mnjuly1970 last updated on 09/Oct/20

	$t h a n k y o u s o m u c h m a s t e r . . .$
	Commented by TANMAY PANACEA last updated on 09/Oct/20

	$m o s t w e l c o m e s i r$
	Answered by 1549442205PVT last updated on 09/Oct/20
	$((x^4 +1)/(x^6 +1))=((ax+b)/(x^2 +1))+((cx^3 +dx^2 +ex+f)/(x^4 −x^2 +1)) ⇔x^4 +1≡(a+c)^5 +(b+d)x^4 +(c−a+e)x^3 +(f−b+d)x^2 +(a+e)x+b+f ⇔ { ((a+c=0)),((b+d=1)),((b+f=1)),((c−a+e=0)),((f−b+d=0)),((a+e=0)) :}⇔ { ((d=f=1/3)),((b=2/3)),((a=c=e=0)) :} ⇒((x^4 +1)/(x^6 +1))=(2/(3(x^2 +1)))+((x^2 +1)/(3(x^4 −x^2 +1)))(1) ((x^2 +1)/(x^4 −x^2 +1))=((ax+b)/(x^2 +x(√3)+1))+((cx+d)/(x^2 −x(√3)+1)) Similarly,we get a=c=0,b=d=1/2 ⇒((x^2 +1)/(x^4 −x^2 +1))=(1/(2(x^2 +x(√3)+1)))+(1/(2(x^2 −x(√3)+1)))(2) From (1)(2)we get Ω=∫_0 ^( 1) ((x^4 +1)/(x^6 +1))dx=(2/3)∫_0 ^( 1) (dx/(x^2 +1)) +(1/6)∫_0 ^( 1) (dx/(x^2 +x(√3)+1))+(1/6)∫_0 ^( 1) (dx/(x^2 −x(√3)+1)) =(2/3)tan^(−1) (x)∣_0 ^1 +(1/6)∫_0 ^( 1) ((d(x+((√3)/2)))/((x+((√3)/2))^2 +((1/2))^2 )) +(1/6)∫_0 ^( 1) ((d(x−((√3)/2)))/((x−((√3)/2))^2 +((1/2))^2 ))(∫(dx/(x^2 +a^2 ))=(1/a)tan^(−1) ((x/a))) =[(2/3)tan^(−1) (x)+(1/3)tan^(−1) (2x+(√3)) +(1/3)tan^(−1) (2x−(√3))]_0 ^1 =(2/3)×(π/4)+(1/3)×((5π)/(12))+(1/3)×(π/(12))=(π/6)+(π/6) Thus,we obtain Ω=(π/3)$
	$\frac{x^{4} + 1}{x^{6} + 1} = \frac{ax + b}{x^{2} + 1} + \frac{{cx}^{3} + {dx}^{2} + ex + f}{x^{4} - x^{2} + 1}$ $\Leftrightarrow x^{4} + 1 \equiv {(a + c)}^{5} + (b + d) x^{4} + (c - a + e) x^{3}$ $+ (f - b + d) x^{2} + (a + e) x + b + f$ $\Leftrightarrow {\begin{cases} a + c = 0 \\ b + d = 1 \\ b + f = 1 \\ c - a + e = 0 \\ f - b + d = 0 \\ a + e = 0 \end{cases} \Leftrightarrow {\begin{cases} d = f = 1 / 3 \\ b = 2 / 3 \\ a = c = e = 0 \end{cases}$ $\Rightarrow \frac{x^{4} + 1}{x^{6} + 1} = \frac{2}{3 (x^{2} + 1)} + \frac{x^{2} + 1}{3 (x^{4} - x^{2} + 1)} (1)$ $\frac{x^{2} + 1}{x^{4} - x^{2} + 1} = \frac{ax + b}{x^{2} + x \sqrt{3} + 1} + \frac{cx + d}{x^{2} - x \sqrt{3} + 1}$ $Similarly, we get a = c = 0, b = d = 1 / 2$ $\Rightarrow \frac{x^{2} + 1}{x^{4} - x^{2} + 1} = \frac{1}{2 (x^{2} + x \sqrt{3} + 1)} + \frac{1}{2 (x^{2} - x \sqrt{3} + 1)} (2)$ $From (1) (2) we get$ $Ω = \int_{0}^{1} \frac{x^{4} + 1}{x^{6} + 1} dx = \frac{2}{3} \int_{0}^{1} \frac{dx}{x^{2} + 1}$ $+ \frac{1}{6} \int_{0}^{1} \frac{dx}{x^{2} + x \sqrt{3} + 1} + \frac{1}{6} \int_{0}^{1} \frac{dx}{x^{2} - x \sqrt{3} + 1}$ $= \frac{2}{3} \tan^{- 1} (x) ∣_{0}^{1} + \frac{1}{6} \int_{0}^{1} \frac{d (x + \frac{\sqrt{3}}{2})}{{(x + \frac{\sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2}}$ $+ \frac{1}{6} \int_{0}^{1} \frac{d (x - \frac{\sqrt{3}}{2})}{{(x - \frac{\sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2}} (\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}))$ $= [\frac{2}{3} \tan^{- 1} (x) + \frac{1}{3} \tan^{- 1} (2 x + \sqrt{3})$ ${+ \frac{1}{3} \tan^{- 1} (2 x - \sqrt{3})]}_{0}^{1}$ $= \frac{2}{3} \times \frac{π}{4} + \frac{1}{3} \times \frac{5 π}{12} + \frac{1}{3} \times \frac{π}{12} = \frac{π}{6} + \frac{π}{6}$ $Thus, we obtain Ω = \frac{π}{3}$
	Commented by mnjuly1970 last updated on 09/Oct/20

	$t h a n k y o u s o m u c h$ $y o u r w o r k i s r e a l l y$ $a d m i r a b l e .$
	Commented by 1549442205PVT last updated on 10/Oct/20

	$Thank Sir . You are welcome .$
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