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Question Number 105231 by mathmax by abdo last updated on 27/Jul/20
calculate ∫_0 ^(+∞) ((2x^2 −1)/((x^2 +x+1)^2 (x^2 −x+1)^2 ))dx
calculate∫0+∞2x2−1(x2+x+1)2(x2−x+1)2dx
Answered by 1549442205PVT last updated on 27/Jul/20
F=(1/2)∫((4x^2 )/((x^2 +x+1)^2 (x^2 −x+1)^2 ))dx−∫(dx/((x^2 +x+1)^2 (x^2 −x+1)^2 ))=∫(A)+∫(B) we have B= (1/((x^2 +x+1)^2 (x^2 −x+1)^2 ))=[(1/2)(((x+1)/((x^2 +x+1)))−((x−1)/((x^2 −x+1))))]^2 =((x^2 +2x+1)/(4(x^2 +x+1)^2 ))+((x^2 −2x+1)/(4(x^2 −x+1)^2 ))−((2x^2 −2)/(4(x^2 +x+1)(x^2 −x+1))) =(1/(4(x^2 +x+1)))+(x/(4(x^2 +x+1)^2 ))+(1/(4(x^2 −x+1)))+((−x)/(4(x^2 −x+1)^2 ))+((2x+1)/(4(x^2 +x+1)))−((2x−1)/(4(x^2 −x+1))) ((2x+2)/(4(x^2 +x+1)))−((2x−2)/(4(x^2 −x+1)))+(x/(4(x^2 +x+1)^2 ))+((−x)/(4(x^2 −x+1)^2 ))(set { ((x^2 +x+1=u)),((x^2 −x+1=v)) :} ) 4B=(x/u^2 )−(x/v^2 )+((2x+2)/u)−((2x−2)/v) A=(1/2)((1/(x^2 −x+1))−(1/(x^2 +x+1)))^2 =((1/((x^2 −x+1)^2 ))+(1/((x^2 +x+1)^2 ))−(2/((x^2 +x+1)(x^2 −x+1))))dx 2A=(1/((x^2 −x+1)^2 ))+(1/((x^2 +x+1)^2 ))+(((x−1)/(x^2 −x+1))−((x+1)/(x^2 +x+1))) =(1/u^2 )+(1/v^2 )+((x−1)/v)−((x+1)/u) ⇒4A=(2/u^2 )+(2/v^2 )+((2x−2)/v)−((2x+2)/u).Hence, 4F=4A−4B=−((x−2)/u^2 )+((x+2)/v^2 )−((4x+1))/u)+((4(x−1))/v) 4F=∫(−((x−2)/((x^2 +x+1)^2 ))+((x+2)/((x^2 −x+1)^2 ))−((4x+1))/(x^2 +x+1))+((4(x−1))/(x^2 −x+1)))dx We have:∫(dx/(x^2 +x+1))=∫((d(x+(1/2)))/((x+(1/2))^2 +(((√3)/2))^2 ))=(2/( (√3)))tan^(−1) (((x+(1/2))/((√3)/2))) =(2/( (√3)))tan^(−1) (((2x+1)/( (√3))))(as ∫(dx/((x^2 +a^2 )))=(1/a)tan^(−1) ((x/a))) Consequently, M=∫((x−2)/((x^2 +x+1)^2 ))dx=(1/2)∫((2x+1)/((x^2 +x+1)^2 ))−(5/2)∫(dx/((x^2 +x+1)^2 )) =(1/2)∫((d(x^2 +x+1))/((x^2 +x+1)^2 ))−(3/2)I_2 =−(1/(2(x^2 +x+1)^2 ))+(1/2)I_2 N=∫((x+2)/((x^2 −x+1)^2 ))dx=(1/2)∫((2x−1)/((x^2 −x+1)^2 ))dx +(5/2)∫(dx/((x^2 −x+1)^2 ))=−(1/(2(x^2 −x+1)^2 ))+(5/2).J_2 P=∫((x+1)/(x^2 +x+1))dx=(1/2)∫((2x+1)/(x^2 +x+1))dx+(1/2)∫(dx/(x^2 +x+1)) =(1/2)ln(x^2 +x+1)+(1/( (√3)))tan^(−1) (((2x+1)/( (√3)))) Q=∫((x−1)/(x^2 −x+1))dx=(1/2)ln(x^2 −x+1)−(1/( (√3)))tan^(−1) (((2x−1)/( (√3)))) We have:I_2 =∫(dt/((t^2 +a^2 )^2 ))(with t=x+(1/2) Apply current formular :I_n =(1/(2a^2 (n−1))).(t/((t^2 +a^2 )^(n−1) ))+(1/a^2 ).((2n−3)/(2n−2)).I_(n−1) we get I_2 =(2/3).((x+(1/2))/(x^2 +x+1))+(4/3).(1/2)∫((d(x+(1/2)))/((x+(1/2))^2 +(((√3)/2))^2 )) =((2x+1)/(3(x^2 +x+1)))+.(4/(3(√3)))tan^(−1) (((2x+1)/( (√3)))).Similarly, J_2 =((2x−1)/(3(x^2 −x+1)))+(4/(3(√3)))tan^(−1) (((2x−1)/( (√3)))) Finally,4F=−M+N−4P+4Q=(1/(2(x^2 +x+1)))−(1/(2(x^2 −x+1))) +(5/2){((2x+1)/(3(x^2 +x+1)))+.(4/(3(√3)))tan^(−1) (((2x+1)/( (√3)))) +[((2x−1)/(3(x^2 −x+1)))+(4/(3(√3)))tan^(−1) (((2x−1)/( (√3))))] } −2ln(x^2 +x+1)+2ln(x^2 −x+1)−(4/( (√3)))tan^(−1) (((2x+1)/( (√3))))−(4/( (√3)))tan^(−1) (((2x−1)/( (√3)))) =((10x^3 +2x)/((x^2 +x+1)(x^2 −x+1)))+2ln((x^2 −x+1)/(x^2 +x+1)) −(2/(3(√3)))[tan^(−1) (((2x−1)/( (√3))))+tan^(−1) (((2x+1)/( (√3))))] It is easy to see that 4F(0)=0 4F(+∞)=−(2/(3(√3)))((π/2)+(π/2))=−((2π)/(3(√3))) Therefore,∫_0 ^(+∞) ((2x^2 −1)/((x^2 +x+1)^2 (x^2 −x+1)^2 ))dx=((−2𝛑)/(3(√3)))
F=12∫4x2(x2+x+1)2(x2−x+1)2dx−∫dx(x2+x+1)2(x2−x+1)2=∫(A)+∫(B)wehaveB=1(x2+x+1)2(x2−x+1)2=[12(x+1(x2+x+1)−x−1(x2−x+1))]2=x2+2x+14(x2+x+1)2+x2−2x+14(x2−x+1)2−2x2−24(x2+x+1)(x2−x+1)=14(x2+x+1)+x4(x2+x+1)2+14(x2−x+1)+−x4(x2−x+1)2+2x+14(x2+x+1)−2x−14(x2−x+1)2x+24(x2+x+1)−2x−24(x2−x+1)+x4(x2+x+1)2+−x4(x2−x+1)2(set{x2+x+1=ux2−x+1=v)4B=xu2−xv2+2x+2u−2x−2vA=12(1x2−x+1−1x2+x+1)2=(1(x2−x+1)2+1(x2+x+1)2−2(x2+x+1)(x2−x+1))dx2A=1(x2−x+1)2+1(x2+x+1)2+(x−1x2−x+1−x+1x2+x+1)=1u2+1v2+x−1v−x+1u⇒4A=2u2+2v2+2x−2v−2x+2u.Hence,4F=4A−4B=−x−2u2+x+2v2−4x+1)u+4(x−1)v4F=∫(−x−2(x2+x+1)2+x+2(x2−x+1)2−4x+1)x2+x+1+4(x−1)x2−x+1)dxWehave:∫dxx2+x+1=∫d(x+12)(x+12)2+(32)2=23tan−1(x+1232)=23tan−1(2x+13)(as∫dx(x2+a2)=1atan−1(xa))Consequently,M=∫x−2(x2+x+1)2dx=12∫2x+1(x2+x+1)2−52∫dx(x2+x+1)2=12∫d(x2+x+1)(x2+x+1)2−32I2=−12(x2+x+1)2+12I2N=∫x+2(x2−x+1)2dx=12∫2x−1(x2−x+1)2dx+52∫dx(x2−x+1)2=−12(x2−x+1)2+52.J2P=∫x+1x2+x+1dx=12∫2x+1x2+x+1dx+12∫dxx2+x+1=12ln(x2+x+1)+13tan−1(2x+13)Q=∫x−1x2−x+1dx=12ln(x2−x+1)−13tan−1(2x−13)Wehave:I2=∫dt(t2+a2)2(witht=x+12Applycurrentformular:In=12a2(n−1).t(t2+a2)n−1+1a2.2n−32n−2.In−1wegetI2=23.x+12x2+x+1+43.12∫d(x+12)(x+12)2+(32)2=2x+13(x2+x+1)+.433tan−1(2x+13).Similarly,J2=2x−13(x2−x+1)+433tan−1(2x−13)Finally,4F=−M+N−4P+4Q=12(x2+x+1)−12(x2−x+1)+52{2x+13(x2+x+1)+.433tan−1(2x+13)+[2x−13(x2−x+1)+433tan−1(2x−13)]}−2ln(x2+x+1)+2ln(x2−x+1)−43tan−1(2x+13)−43tan−1(2x−13)=10x3+2x(x2+x+1)(x2−x+1)+2lnx2−x+1x2+x+1−233[tan−1(2x−13)+tan−1(2x+13)]Itiseasytoseethat4F(0)=04F(+∞)=−233(π2+π2)=−2π33Therefore,∫0+∞2x2−1(x2+x+1)2(x2−x+1)2dx=−2π33
Commented by abdomathmax last updated on 27/Jul/20
thank you sir.
thankyousir.

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