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Question Number 187855 by Michaelfaraday last updated on 23/Feb/23

solve ∫((x^4 +x^2 +1)/(2(x^2 +1)))dx

$${solve} \\ $$$$\int\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$

Answered by cortano12 last updated on 23/Feb/23

I=(1/2)∫ (((x^2 +1)^2 −x^2 )/(x^2 +1)) dx I=(1/6)x^3 +(1/2)x−(1/2)∫ (x^2 /(x^2 +1)) dx = (1/6)x^3 +(1/2)x−(1/2)∫ ((tan^2 θ)/(sec^2 θ)) sec^2 θ dθ ; x=tan θ = (1/6)x^3 +(1/2)x−(1/2)∫(sec^2 θ−1)dθ = (1/6)x^3 +(1/2)x−(1/2)x+(1/2)arctan x+C = (1/6)x^3 +(1/2)arctan x+C

$$\:\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx} \\ $$$$\:\mathrm{I}=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{sec}\:^{\mathrm{2}} \theta}\:\mathrm{sec}\:^{\mathrm{2}} \theta\:\mathrm{d}\theta\:;\:\mathrm{x}=\mathrm{tan}\:\theta \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{1}\right)\mathrm{d}\theta \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\mathrm{x}+\mathrm{C} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\mathrm{x}+\mathrm{C}\: \\ $$

Commented by Michaelfaraday last updated on 25/Feb/23

thanks sir

$${thanks}\:{sir} \\ $$

Answered by CElcedricjunior last updated on 23/Feb/23

∫((x^4 +x^2 +1)/(2(x^2 +1)))dx=(1/2)∫[((x^2 (x^2 +1))/((x^2 +1)))+(1/((x^2 +1)))]dx =(1/2)∫[x^2 +(1/(x^2 +1))]dx ■Moivre =(1/2)[(1/3)x^3 +arctan(x)]+k ★Cedric junior

$$\int\frac{\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{\boldsymbol{{x}}^{\mathrm{2}} \left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}\right]\boldsymbol{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\right]\boldsymbol{{dx}}\:\blacksquare{Moivre} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)\right]+\boldsymbol{{k}}\:\bigstar\mathscr{C}{edric}\:{junior} \\ $$

Commented by Michaelfaraday last updated on 25/Feb/23

thanks sir

$${thanks}\:{sir} \\ $$

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