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x-2-1-x-4-x-2-1-dx-




Question Number 135673 by liberty last updated on 14/Mar/21
∫ ((x^2 +1)/(x^4 +x^2 +1)) dx
$$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\: \\ $$
Answered by Olaf last updated on 15/Mar/21
Ω = ∫((x^2 +1)/(x^4 +x^2 +1)) dx  Ω = ∫((x^2 +1)/((x^2 −1)^2 +3x^2 )) dx  Ω = ∫(((x^2 +1)/(((√(3x)))^2 ))/(1+(((x^2 −1)/( (√3)x)))^2 )) dx  Ω = (1/( (√3)))∫((((2x)((√3)x)−(x^2 −1)(√3))/(((√(3x)))^2 ))/(1+(((x^2 −1)/( (√3)x)))^2 )) dx  Ω = (1/( (√3)))∫((d(((x^2 −1)/( (√3)x))))/(1+(((x^2 −1)/( (√3)x)))^2 ))   Ω = (1/( (√3)))arctan(((x^2 −1)/( (√3)x)))+C
$$\Omega\:=\:\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$\Omega\:=\:\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} }\:{dx} \\ $$$$\Omega\:=\:\int\frac{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\left(\sqrt{\mathrm{3}{x}}\right)^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{\frac{\left(\mathrm{2}{x}\right)\left(\sqrt{\mathrm{3}}{x}\right)−\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{3}}}{\left(\sqrt{\mathrm{3}{x}}\right)^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{{d}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)}{\mathrm{1}+\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)^{\mathrm{2}} }\: \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{x}}\right)+\mathrm{C} \\ $$
Commented by liberty last updated on 15/Mar/21
yes..thank you
$${yes}..{thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 15/Mar/21
I =∫ ((x^2  +1)/(x^4  +x^2  +1))dx ⇒I =∫ ((1+(1/x^2 ))/(x^2  +1+(1/x^2 )))dx =∫ ((1+(1/x^2 ))/((x−(1/x))^2 +3))dx  =_(x−(1/x)=t)    ∫  (dt/(t^2  +3)) =_(t=(√3)y)    ∫  (((√3)dy)/(3y^2  +3)) =(1/( (√3)))∫ (dy/(1+y^2 ))  =(1/( (√3)))arctany +C =(1/( (√3)))arctan((1/( (√3)))(x−(1/x))) +C
$$\mathrm{I}\:=\int\:\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{3}}\mathrm{dx} \\ $$$$=_{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t}} \:\:\:\int\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}}\:=_{\mathrm{t}=\sqrt{\mathrm{3}}\mathrm{y}} \:\:\:\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{dy}}{\mathrm{3y}^{\mathrm{2}} \:+\mathrm{3}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\:\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctany}\:+\mathrm{C}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)\right)\:+\mathrm{C} \\ $$

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