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Question Number 28701 by students last updated on 29/Jan/18
integration (x^3 /(x^2 +x+1))
$${integration}\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$
Commented by abdo imad last updated on 29/Jan/18
∫ (x^3 /(x^2 +x+1))dx= ∫ ((x^3 −1 +1)/(x^2 +x+1))dx= ∫(x−1)dx ++∫ (dx/(x^2 +x+1)) = (x^2 /2) −x + ∫ (dx/(x^2 +x+1)) but I=∫ (dx/(x^2 +x+1)) =∫ (dx/((x+(1/2))^2 +(3/4))) let use the ch.x+(1/2)=((√3)/2)t I= ∫ (1/((3/4)t^2 +(3/4))) ((√3)/2)dt=((√3)/2)(4/3) ∫ (dt/(1+t^2 ))=((2(√3))/3) arctant = ((2(√3))/3) arctan( (2/( (√3)))(x+(1/2))) so ∫ (x^3 /(x^2 +x+1))dx= (x^2 /2)−x +((2(√3))/3) arctan((2/( (√3)))(x+(1/2))).
$$\int\:\:\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:\int\:\frac{{x}^{\mathrm{3}} −\mathrm{1}\:+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:\int\left({x}−\mathrm{1}\right){dx}\:++\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−{x}\:+\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\:{but} \\ $$$${I}=\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:=\int\:\:\:\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:\:{let}\:{use}\:{the}\:{ch}.{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t} \\ $$$${I}=\:\int\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\frac{\mathrm{4}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctant} \\ $$$$=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:{so} \\ $$$$\int\:\:\:\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right). \\ $$$$ \\ $$
Commented by abdo imad last updated on 29/Jan/18
∫(...)dx=........+λ.
$$\int\left(…\right){dx}=……..+\lambda. \\ $$
Answered by mrW2 last updated on 29/Jan/18
=((x^3 +x^2 +x−x^2 −x−1+1)/(x^2 +x+1)) =x−1+(1/(x^2 +x+1)) =x−1+(1/((x+(1/2))^2 +(((√3)/2))^2 )) ∫(x^3 /(x^2 +x+1))dx =(x^2 /2)−x+(2/( (√3))) tan^(−1) ((x+(1/2))/((√3)/2))+C =(x^2 /2)−x+(2/( (√3))) tan^(−1) ((2x+1)/( (√3)))+C
$$=\frac{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}−{x}^{\mathrm{2}} −{x}−\mathrm{1}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$={x}−\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$={x}−\mathrm{1}+\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\int\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}+{C} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$

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