calculate ∫_0 ^(+∞) ((2x^2 −1)/((x^2 +x+1)^2 (x^2 −x+1)^2 ))dx


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Question Number 105231 by mathmax by abdo last updated on 27/Jul/20

$calculate \int_{0}^{+ \infty} \frac{{2 x}^{2} - 1}{{(x^{2} + x + 1)}^{2} {(x^{2} - x + 1)}^{2}} dx$
	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)))$
	$F = \frac{1}{2} \int \frac{{4 x}^{2}}{{(x^{2} + x + 1)}^{2} {(x^{2} - x + 1)}^{2}} dx - \int \frac{dx}{{(x^{2} + x + 1)}^{2} {(x^{2} - x + 1)}^{2}} = \int (A) + \int (B)$ $we have B = \frac{1}{{(x^{2} + x + 1)}^{2} {(x^{2} - x + 1)}^{2}} = {[\frac{1}{2} (\frac{x + 1}{(x^{2} + x + 1)} - \frac{x - 1}{(x^{2} - x + 1)})]}^{2} = \frac{x^{2} + 2 x + 1}{4 {(x^{2} + x + 1)}^{2}} + \frac{x^{2} - 2 x + 1}{4 {(x^{2} - x + 1)}^{2}} - \frac{{2 x}^{2} - 2}{4 (x^{2} + x + 1) (x^{2} - x + 1)}$ $= \frac{1}{4 (x^{2} + x + 1)} + \frac{x}{4 {(x^{2} + x + 1)}^{2}} + \frac{1}{4 (x^{2} - x + 1)} + \frac{- x}{4 {(x^{2} - x + 1)}^{2}} + \frac{2 x + 1}{4 (x^{2} + x + 1)} - \frac{2 x - 1}{4 (x^{2} - x + 1)}$ $\frac{2 x + 2}{4 (x^{2} + x + 1)} - \frac{2 x - 2}{4 (x^{2} - x + 1)} + \frac{x}{4 {(x^{2} + x + 1)}^{2}} + \frac{- x}{4 {(x^{2} - x + 1)}^{2}} (set {\begin{cases} x^{2} + x + 1 = u \\ x^{2} - x + 1 = v \end{cases})$ $4 B = \frac{x}{u^{2}} - \frac{x}{v^{2}} + \frac{2 x + 2}{u} - \frac{2 x - 2}{v}$ $A = \frac{1}{2} {(\frac{1}{x^{2} - x + 1} - \frac{1}{x^{2} + x + 1})}^{2} = (\frac{1}{{(x^{2} - x + 1)}^{2}} + \frac{1}{{(x^{2} + x + 1)}^{2}} - \frac{2}{(x^{2} + x + 1) (x^{2} - x + 1)}) dx$ $2 A = \frac{1}{{(x^{2} - x + 1)}^{2}} + \frac{1}{{(x^{2} + x + 1)}^{2}} + (\frac{x - 1}{x^{2} - x + 1} - \frac{x + 1}{x^{2} + x + 1})$ $= \frac{1}{u^{2}} + \frac{1}{v^{2}} + \frac{x - 1}{v} - \frac{x + 1}{u}$ $\Rightarrow 4 A = \frac{2}{u^{2}} + \frac{2}{v^{2}} + \frac{2 x - 2}{v} - \frac{2 x + 2}{u} . Hence,$ $4 F = 4 A - 4 B = - \frac{x - 2}{u^{2}} + \frac{x + 2}{v^{2}} - \frac{4 x + 1)}{u} + \frac{4 (x - 1)}{v}$ $4 F = \int (- \frac{x - 2}{{(x^{2} + x + 1)}^{2}} + \frac{x + 2}{{(x^{2} - x + 1)}^{2}} - \frac{4 x + 1)}{x^{2} + x + 1} + \frac{4 (x - 1)}{x^{2} - x + 1}) dx$ $We have : \int \frac{dx}{x^{2} + x + 1} = \int \frac{d (x + \frac{1}{2})}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}})$ $= \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) (as \int \frac{dx}{(x^{2} + a^{2})} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}))$ $Consequently,$ $M = \int \frac{x - 2}{{(x^{2} + x + 1)}^{2}} dx = \frac{1}{2} \int \frac{2 x + 1}{{(x^{2} + x + 1)}^{2}} - \frac{5}{2} \int \frac{dx}{{(x^{2} + x + 1)}^{2}}$ $= \frac{1}{2} \int \frac{d (x^{2} + x + 1)}{{(x^{2} + x + 1)}^{2}} - \frac{3}{2} I_{2} = - \frac{1}{2 {(x^{2} + x + 1)}^{2}} + \frac{1}{2} I_{2}$ $N = \int \frac{x + 2}{{(x^{2} - x + 1)}^{2}} dx = \frac{1}{2} \int \frac{2 x - 1}{{(x^{2} - x + 1)}^{2}} dx$ $+ \frac{5}{2} \int \frac{dx}{{(x^{2} - x + 1)}^{2}} = - \frac{1}{2 {(x^{2} - x + 1)}^{2}} + \frac{5}{2} . J_{2}$ $P = \int \frac{x + 1}{x^{2} + x + 1} dx = \frac{1}{2} \int \frac{2 x + 1}{x^{2} + x + 1} dx + \frac{1}{2} \int \frac{dx}{x^{2} + x + 1}$ $= \frac{1}{2} \ln (x^{2} + x + 1) + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}})$ $Q = \int \frac{x - 1}{x^{2} - x + 1} dx = \frac{1}{2} \ln (x^{2} - x + 1) - \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}})$ $We have : I_{2} = \int \frac{dt}{{(t^{2} + a^{2})}^{2}} (with t = x + \frac{1}{2}$ $Apply current formular : I_{n} = \frac{1}{{2 a}^{2} (n - 1)} . \frac{t}{{(t^{2} + a^{2})}^{n - 1}} + \frac{1}{a^{2}} . \frac{2 n - 3}{2 n - 2} . I_{n - 1} we get$ $I_{2} = \frac{2}{3} . \frac{x + \frac{1}{2}}{x^{2} + x + 1} + \frac{4}{3} . \frac{1}{2} \int \frac{d (x + \frac{1}{2})}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$ $= \frac{2 x + 1}{3 (x^{2} + x + 1)} + . \frac{4}{3 \sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) . Similarly,$ $J_{2} = \frac{2 x - 1}{3 (x^{2} - x + 1)} + \frac{4}{3 \sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}})$ $Finally, 4 F = - M + N - 4 P + 4 Q = \frac{1}{2 (x^{2} + x + 1)} - \frac{1}{2 (x^{2} - x + 1)}$ $+ \frac{5}{2} {\frac{2 x + 1}{3 (x^{2} + x + 1)} + . \frac{4}{3 \sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}})$ $+ [\frac{2 x - 1}{3 (x^{2} - x + 1)} + \frac{4}{3 \sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}})]}$ $- 2 \ln (x^{2} + x + 1) + 2 \ln (x^{2} - x + 1) - \frac{4}{\sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) - \frac{4}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}})$ $= \frac{{10 x}^{3} + 2 x}{(x^{2} + x + 1) (x^{2} - x + 1)} + 2 \ln \frac{x^{2} - x + 1}{x^{2} + x + 1}$ $- \frac{2}{3 \sqrt{3}} [\tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}})]$ $It is easy to see that 4 F (0) = 0$ $4 F (+ \infty) = - \frac{2}{3 \sqrt{3}} (\frac{π}{2} + \frac{π}{2}) = - \frac{2 π}{3 \sqrt{3}}$ $Th erefore, \int_{0}^{+ \infty} \frac{2 x^{2} - 1}{{(x^{2} + x + 1)}^{2} {(x^{2} - x + 1)}^{2}} dx = \frac{- 2 π}{3 \sqrt{3}}$
	Commented by abdomathmax last updated on 27/Jul/20

	$thank you sir .$
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