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

         ...  nice  calculus...   please  evaluate ::          Ω=∫_0 ^( 1) ((x^4 +1)/(x^6 +1))dx =???           m.n.1970

...nicecalculus...pleaseevaluate::Ω=01x4+1x6+1dx=???m.n.1970

Answered by TANMAY PANACEA last updated on 09/Oct/20

∫(((x^2 +1)^2 −2x^2 )/((x^2 +1)(x^4 −x^2 +1)))dx  ∫((x^2 +1)/(x^4 −x^2 +1))dx−∫((2x^2 )/((x^3 )^2 +1))dx  ∫((1+(1/x^2 ))/(x^2 +(1/x^2 )−1))dx−(2/3)∫((d(x^3 ))/((x^3 )^2 +1))  ∫((d(x−(1/x)))/((x−(1/x))^2 +1))−(2/3)∫((d(x^3 ))/((x^3 )^2 +1))  tan^(−1) (x−(1/x))−(2/3)tan^(−1) (x^3 )+C  now put limit  ∣tan^(−1) (x−(1/x))−(2/3)tan^(−1) (x^3 )∣_0 ^1   tan^(−1) (0)−(2/3)tan^(−1) (1)−tan^(−1) (0−∞)+(2/3)tan^(−1) (0)  =(π/4)×((−2)/3)+(π/2)  =(π/2)(1−(1/3))=(π/3)

(x2+1)22x2(x2+1)(x4x2+1)dxx2+1x4x2+1dx2x2(x3)2+1dx1+1x2x2+1x21dx23d(x3)(x3)2+1d(x1x)(x1x)2+123d(x3)(x3)2+1tan1(x1x)23tan1(x3)+Cnowputlimittan1(x1x)23tan1(x3)01tan1(0)23tan1(1)tan1(0)+23tan1(0)=π4×23+π2=π2(113)=π3

Commented by mnjuly1970 last updated on 09/Oct/20

thank you so much  master...

thankyousomuchmaster...

Commented by TANMAY PANACEA last updated on 09/Oct/20

most welcome sir

mostwelcomesir

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)

x4+1x6+1=ax+bx2+1+cx3+dx2+ex+fx4x2+1x4+1(a+c)5+(b+d)x4+(ca+e)x3+(fb+d)x2+(a+e)x+b+f{a+c=0b+d=1b+f=1ca+e=0fb+d=0a+e=0{d=f=1/3b=2/3a=c=e=0x4+1x6+1=23(x2+1)+x2+13(x4x2+1)(1)x2+1x4x2+1=ax+bx2+x3+1+cx+dx2x3+1Similarly,wegeta=c=0,b=d=1/2x2+1x4x2+1=12(x2+x3+1)+12(x2x3+1)(2)From(1)(2)wegetΩ=01x4+1x6+1dx=2301dxx2+1+1601dxx2+x3+1+1601dxx2x3+1=23tan1(x)01+1601d(x+32)(x+32)2+(12)2+1601d(x32)(x32)2+(12)2(dxx2+a2=1atan1(xa))=[23tan1(x)+13tan1(2x+3)+13tan1(2x3)]01=23×π4+13×5π12+13×π12=π6+π6Thus,weobtainΩ=π3

Commented by mnjuly1970 last updated on 09/Oct/20

thank you so much  your work is really  admirable.

thankyousomuchyourworkisreallyadmirable.

Commented by 1549442205PVT last updated on 10/Oct/20

Thank Sir.You are welcome.

ThankSir.Youarewelcome.

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