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Question Number 66326 by mathmax by abdo last updated on 12/Aug/19
calculate ∫_0 ^1 ((x^4 +1)/(x^6 +1))dx
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx} \\ $$
Commented by Prithwish sen last updated on 13/Aug/19
((x^4 +1)/(x^6 +1)) = (A/(x^2 +1)) + (B/(x^2 +(√3)x+1)) +(C/(x^2 −(√3)x+1)) x^4 +1= x^4 (A+B+C)+x^3 ((√3)C−(√3)B)+x^2 (2B+2C−A)+x((√3)C−(√3)B)+(A+B+C) ⇒A+B+C = 1 (√3)C−(√3)B=0 2B+2C−A=0 A+B+C=0 A=(2/3) B=(1/6) C=(1/6) ∫((2dx)/(3(1+x^2 ))) +(1/6)∫(dx/((x+((√3)/2))^2 +((1/2))^2 )) +(1/6) ∫(dx/((x−((√3)/2))^2 +((1/2))^2 ))
$$\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\:=\:\frac{\mathrm{A}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{B}}{\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}\mathrm{x}+\mathrm{1}}\:+\frac{\mathrm{C}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}\mathrm{x}+\mathrm{1}} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{1}=\:\mathrm{x}^{\mathrm{4}} \left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+\mathrm{x}^{\mathrm{3}} \left(\sqrt{\mathrm{3}}\mathrm{C}−\sqrt{\mathrm{3}}\mathrm{B}\right)+\mathrm{x}^{\mathrm{2}} \left(\mathrm{2B}+\mathrm{2C}−\mathrm{A}\right)+\mathrm{x}\left(\sqrt{\mathrm{3}}\mathrm{C}−\sqrt{\mathrm{3}}\mathrm{B}\right)+\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right) \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\mathrm{1} \\ $$$$\sqrt{\mathrm{3}}\mathrm{C}−\sqrt{\mathrm{3}}\mathrm{B}=\mathrm{0} \\ $$$$\mathrm{2B}+\mathrm{2C}−\mathrm{A}=\mathrm{0} \\ $$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{0} \\ $$$$\mathrm{A}=\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{B}=\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{C}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\int\frac{\mathrm{2dx}}{\mathrm{3}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\:+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{6}}\:\int\frac{\mathrm{dx}}{\left(\mathrm{x}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 13/Aug/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Commented by Prithwish sen last updated on 13/Aug/19
welome
$$\mathrm{welome} \\ $$
Commented by mathmax by abdo last updated on 13/Aug/19
∫_0 ^(+∞) ((x^4 +1)/(x^6 +1))dx =∫_0 ^1 ((x^4 +1)/(x^6 +1))dx +∫_1 ^(+∞) ((x^4 +1)/(x^6 +1))dx but ∫_1 ^(+∞) ((x^4 +1)/(x^6 +1))dx =_(x=(1/t)) −∫_0 ^1 (((1/t^4 )+1)/((1/t^6 )+1))(−(dt/t^2 )) =∫_0 ^1 ((1+t^4 )/(1+t^6 ))dt ⇒ ∫_0 ^∞ ((x^4 +1)/(x^6 +1))dx =2 ∫_0 ^1 ((x^4 +1)/(x^6 +1))dx ⇒∫_0 ^1 ((x^4 +1)/(x^6 +1))dx=(1/2)∫_0 ^∞ (x^4 /(1+x^6 ))dx +(1/2)∫_0 ^∞ (dx/(1+x^6 )) changement x =t^(1/6) give ∫_0 ^∞ (x^4 /(1+x^6 ))dx =∫_0 ^∞ (t^(2/3) /(1+t))(1/6)t^((1/6)−1) dt =(1/6)∫_0 ^∞ (t^((2/3)+(1/6)−1) /(1+t))dt =(1/6)∫_0 ^∞ (t^((5/6)−1) /(1+t))dt =(1/6)(π/(sin(((5π)/6)))) =(π/6)×2=(π/3) ∫_0 ^∞ (dx/(1+x^6 )) =_(x=t^(1/6) ) (1/6)∫_0 ^∞ (t^((1/6)−1) /(1+t))dt=(1/6)(π/(sin((π/6)))) =(π/6)×2 =(π/3) ⇒ ∫_0 ^1 ((x^4 +1)/(x^6 +1))dx =(π/6)+(π/6) =(π/3)
$$\int_{\mathrm{0}} ^{+\infty} \:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:{but} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\frac{\mathrm{1}}{{t}^{\mathrm{4}} }+\mathrm{1}}{\frac{\mathrm{1}}{{t}^{\mathrm{6}} }+\mathrm{1}}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{t}^{\mathrm{4}} }{\mathrm{1}+{t}^{\mathrm{6}} }{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} } \\ $$$${changement}\:{x}\:={t}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}+{t}}\frac{\mathrm{1}}{\mathrm{6}}{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{5}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{1}}{\mathrm{6}}\frac{\pi}{{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{6}}×\mathrm{2}=\frac{\pi}{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} }\:=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{6}}} } \:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\mathrm{1}}{\mathrm{6}}\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{6}}×\mathrm{2}\:=\frac{\pi}{\mathrm{3}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{1}}{{x}^{\mathrm{6}} \:+\mathrm{1}}{dx}\:=\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}} \\ $$

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