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Question Number 86824 by Ar Brandon last updated on 31/Mar/20
∫((x^6 +x^2 )/(x^8 −x^4 +1))dx
$$\int\frac{{x}^{\mathrm{6}} +{x}^{\mathrm{2}} }{{x}^{\mathrm{8}} −{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$
Answered by TANMAY PANACEA. last updated on 31/Mar/20
∫((x^2 +(1/x^2 ))/(x^4 −1+(1/x^4 )))dx ∫((x^2 +(1/x^2 ))/((x^2 +(1/x^2 ))^2 −3))dx (1/2)∫((x^2 +(1/x^2 )+(√3) +x^2 +(1/x^2 )−(√3))/((x^2 +(1/x^2 )+(√3) )(x^2 +(1/x^2 )−(√3) )))dx (1/2)∫(dx/(x^2 +(1/x^2 )−(√3)))+(1/2)∫(dx/(x^2 +(1/x^2 )+(√3))) (1/4)∫((1−(1/x^2 )+1+(1/x^2 ))/(x^2 +(1/x^2 )−(√3)))dx+(1/4)∫((1−(1/x^2 )+1+(1/x^2 ))/(x^2 +(1/x^2 )+(√3)))dx (1/4)∫((d(x+(1/x)))/((x+(1/x))^2 −(2+(√3) )))+(1/4)∫((d(x−(1/x)))/((x−(1/x))^2 +2−(√3) ))+(1/4)∫((d(x+(1/x)))/((x+(1/x))^2 −2+(√3)))+(1/4)∫((d(x−(1/x)))/((x−(1/x))^2 +2+(√3))) (1/4)×(1/(2(√(2+(√3) ))))ln(((x+(1/x)−(√(2+(√3))))/(x+(1/x)+(√(2+(√3))))))=I_1 (1/4)×(1/( (√(2−(√3)))))tan^(−1) (((x−(1/x))/( (√(2−(√3))))))=I_2 (1/4)×(1/(2(√(2−(√3)))))ln(((x+(1/x)−(√(2−(√3))))/(x+(1/x)+(√(2−(√3))))))=I_3 (1/4)×(1/( (√(2+(√3)))))tan^(−1) (((x−(1/x))/( (√(2+(√3)))))) pls add them...
$$\int\frac{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{4}} −\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{3}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}}\:+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{3}}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}}\:\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{3}}\:\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{3}}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}−\sqrt{\mathrm{3}}\:}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}\:}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\right)={I}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}\right)={I}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}\right)={I}_{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\right) \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{add}}\:\boldsymbol{{them}}… \\ $$
Commented by Ar Brandon last updated on 02/Apr/20
Amazing !
$${Amazing}\:\:! \\ $$
Commented by TANMAY PANACEA. last updated on 02/Apr/20
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Answered by M±th+et£s last updated on 03/Apr/20
(1/2)∫((2(x^6 +x^2 ))/(x^8 +2x^4 +1−3x^4 ))dx=(1/2)∫((x^6 +x^6 +(√3)x^4 −(√3)x^4 +x^2 +x^2 )/((x^4 +1)^2 −3x^4 ))dx (1/2)∫((x^6 +(√3)x^4 +x^2 +x^6 −(√3)x^4 +x^2 )/((x^4 +1−(√3)x^2 )(x^4 +1+(√3)x^2 )))dx (1/2)∫((x^2 (x^4 +(√3)x^2 +1)+x^2 (x^4 −(√3)x^2 +1))/((x^4 +1−(√3)x^2 )(x^4 +1+(√3)x^2 )))dx (1/4)∫(((2x^2 )/((x^4 +1+(√3)x^2 ))) + ((2x^2 )/((x^4 +1−(√3)x^2 )))dx (1/4)∫(((x^2 +1+x^2 −1)/((x^4 +1+(√3)x^2 )))+((x^2 +1+x^2 −1)/((x^4 +1−(√3)x^2 ))))dx (1/4)∫(((x^2 +1)/(x^4 +1+(√3)x^2 )) +((x^2 −1)/(x^4 +1+(√3)x^2 ))+((x^2 +1)/(x^4 +1−(√3)x^2 ))+((x^2 −1)/(x^4 +1−(√3)x^2 )))dx (1/4)∫(((1+1/x^2 )/(x^2 +(√3)+1/x^2 ))dx +((1−1/x^2 )/(x^2 +(√3)+1/x^2 ))+((1+1/x^2 )/(x^2 −(√3)+1/x^2 ))+((1−1/x^2 )/(x^2 −(√3)+1/x^2 )))dx (1/4)∫(((1+1/x^2 )/(x^2 −2+1/x^2 −(√3)+2))+((1−1/x^2 )/(x^2 +2+1/x^2 −2−(√3)))+((1+1/x^2 )/(x^2 −2+1/x^2 +2+(√3)))+((1−1/x^2 )/(x^2 +2+1/x^2 −2+(√3)))) (1/4)∫(((1+1/x^2 )/((x−1/x)^2 −((√3)−2)))+((1−1/x^2 )/((x+1/x)^2 −(2+(√3))))+((1+1/x^2 )/((x−1/x)^2 +((√3)+2)))+((1−1/x^2 )/((x−1/x)^2 +((√3)−2))))dx (1/(4(√(2−(√3)))))tanh^(−1) (1/( (√(2−(√3)))))(x−(1/x))+(1/(4(√(2−(√3)))))tanh^(−1) (1/( (√(2−(√3)))))(x+(1/x))+(1/(4(√((√3)+2))))tan^(−1) (1/( (√((√3)+2))))(x−(1/x))+(1/(4(√((√3)−2))))tan^(−1) (1/( (√((√3)−2))))(x+(1/x))+c pls check
$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}\left({x}^{\mathrm{6}} +{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{1}−\mathrm{3}{x}^{\mathrm{4}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{6}} +{x}^{\mathrm{6}} +\sqrt{\mathrm{3}}{x}^{\mathrm{4}} −\sqrt{\mathrm{3}}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{4}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{6}} +\sqrt{\mathrm{3}}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:+{x}^{\mathrm{6}} −\sqrt{\mathrm{3}}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} +\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{1}\right)+{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right.}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)}+\frac{{x}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \right)}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}+\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} }\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}+\mathrm{1}/{x}^{\mathrm{2}} }{dx}\:+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}+\mathrm{1}/{x}^{\mathrm{2}} }+\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}+\mathrm{1}/{x}^{\mathrm{2}} }+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}+\mathrm{1}/{x}^{\mathrm{2}} }\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}+\mathrm{1}/{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}+\mathrm{2}}+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{2}+\mathrm{1}/{x}^{\mathrm{2}} −\mathrm{2}−\sqrt{\mathrm{3}}}+\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}+\mathrm{1}/{x}^{\mathrm{2}} +\mathrm{2}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{2}+\mathrm{1}/{x}^{\mathrm{2}} −\mathrm{2}+\sqrt{\mathrm{3}}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}/{x}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}−\mathrm{2}\right)}+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{\left({x}+\mathrm{1}/{x}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}+\frac{\mathrm{1}+\mathrm{1}/{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}/{x}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)}+\frac{\mathrm{1}−\mathrm{1}/{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}/{x}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}−\mathrm{2}\right)}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{tanh}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}\left({x}−\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{tanh}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\sqrt{\mathrm{3}}+\mathrm{2}}}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\mathrm{2}}}\left({x}−\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\sqrt{\mathrm{3}}−\mathrm{2}}}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}−\mathrm{2}}}\left({x}+\frac{\mathrm{1}}{{x}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${pls}\:{check} \\ $$$$ \\ $$

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