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Question Number 181319 by a.lgnaoui last updated on 23/Nov/22

Determiner  1.   ∫(x/(x^4 +x^2 +1))dx  2.   ∫((x^4 +1)/(x^4 +x^2 +1))dx

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

Answered by floor(10²Eta[1]) last updated on 23/Nov/22

∫(x/(x^4 +x^2 +1))dx=I  u=x^2 ⇒(du/2)=xdx  I=(1/2)∫(du/(u^2 +u+1))=(2/3)∫(du/((((2u)/( (√3)))+(1/( (√3))))^2 +1))  t=((2u+1)/( (√3)))⇒dt=(2/( (√3)))du  I=((√3)/3)∫(dt/(t^2 +1))=((√3)/3)arctg(t)+C  ⇒I=((√3)/3)arctg(((2x^2 +1)/( (√3))))+C

$$\int\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}=\mathrm{I} \\ $$$$\mathrm{u}=\mathrm{x}^{\mathrm{2}} \Rightarrow\frac{\mathrm{du}}{\mathrm{2}}=\mathrm{xdx} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{du}}{\left(\frac{\mathrm{2u}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{t}=\frac{\mathrm{2u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\mathrm{dt}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{du} \\ $$$$\mathrm{I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctg}\left(\mathrm{t}\right)+\mathrm{C} \\ $$$$\Rightarrow\mathrm{I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctg}\left(\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\mathrm{C} \\ $$

Commented by a.lgnaoui last updated on 24/Nov/22

thanks

$${thanks}\: \\ $$

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