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Question Number 34225 by abdo imad last updated on 03/May/18

find ∫ (dx/(1+x^2 +x^4 ))

$${find}\:\int\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} } \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18

∫(1/(x^4 +x^(2 ) +1))dx ∫((1/x^2 )/(x^2 +(1/(x^2 ))+1)) I.=(1/2)∫(((1+(1/(x^2 )))−(1−(1/x^2 ) ))/(x^2 +(1/(x^2 ))+1))dx I_(1=) (1/2)∫(((1+(1/(x^2 ))))/((x−(1/x))^2 +3))dx =(1/2)∫(dt_1 /(t_1 ^2 +3)) use formula =(1/2).(1/((√3) )).tan^(−1) ((t_1 /(√3))) I_2 =(1/2)∫(((1−(1/x^(2 ) )))/((x+(1/x))^2 −1)) I_2 =(1/2).∫(dt_2 /(t_2 ^2 −1)) use formula I_2 =(1/2).(1/2).ln∣((1−t_2 )/(1+t_2 ))∣ I=I_1 −I_2

$$\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}\:} +\mathrm{1}}{dx} \\ $$$$\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}+\mathrm{1}} \\ $$$${I}.=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}+\mathrm{1}}{dx} \\ $$$${I}_{\mathrm{1}=} \frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}_{\mathrm{1}} }{{t}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{3}}\:{use}\:{formula} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\sqrt{\mathrm{3}}\:}.{tan}^{−\mathrm{1}} \left(\frac{{t}_{\mathrm{1}} }{\sqrt{\mathrm{3}}}\right) \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} }\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}.\int\frac{{dt}_{\mathrm{2}} }{{t}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}}\:{use}\:{formula} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.{ln}\mid\frac{\mathrm{1}−{t}_{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{2}} }\mid \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$

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