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Question Number 91308 by jagoll last updated on 29/Apr/20

∫ (x^2 /(x^4 −x^2 −2)) dx ?

$$\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}\:{dx}\:? \\ $$

Commented by jagoll last updated on 29/Apr/20

∫ (x^2 /((x^2 −2)(x^2 +1))) dx = ∫ (1/(3(x^2 +1)))+(2/(3(x^2 −2))) dx = (1/3)tan^(−1) (x)+ (2/3)∫ (dx/((x+(√2))(x−(√2)))) = (1/3)tan^(−1) (x) +(1/(3(√2)))∫ (1/(x−(√2)))−(1/(x+(√2))) dx (1/3)tan^(−1) (x)+(1/(3(√2))) ln (((x−(√2))/(x+(√2)))) + c

$$\int\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx}\:= \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}}{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\:{dx}\:=\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{2}}\right)\left({x}−\sqrt{\mathrm{2}}\right)}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{2}}}\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\frac{{x}−\sqrt{\mathrm{2}}}{{x}+\sqrt{\mathrm{2}}}\right)\:+\:{c} \\ $$

Commented by mathmax by abdo last updated on 29/Apr/20

let I =∫ (x^2 /(x^4 −x^2 −2)) x^4 −x^2 −2=x^4 +x^2 −2x^2 −2 =x^2 (x^2 +1)−2(x^2 +1)=(x^2 +1)(x^2 −2) ⇒ I =∫ (x^2 /((x^2 +1)(x^2 −2))) =(1/3)∫ x^2 ((1/(x^2 −2))−(1/(x^2 +1)))dx =(1/3)∫ (x^2 /(x^2 −2))dx−(1/3)∫(x^2 /(x^2 +1))dx =(1/3)∫((x^2 −2+2)/(x^2 −2))dx−(1/3)∫((x^2 +1−1)/(x^2 +1))dx =(2/3) ∫ (dx/(x^2 −2)) +(1/3)∫ (dx/(x^2 +1)) =(2/(3×2(√2)))∫((1/(x−(√2)))−(1/(x+(√2))))dx+(1/3)arctanx =(1/(3(√2)))ln∣((x−(√2))/(x+(√2)))∣+(1/3)arctanx +C

$${let}\:{I}\:=\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}={x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\:={x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right) \\ $$$$\Rightarrow\:{I}\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{3}}\int\:{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{2}}{\mathrm{3}×\mathrm{2}\sqrt{\mathrm{2}}}\int\left(\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{2}}}\right){dx}+\frac{\mathrm{1}}{\mathrm{3}}{arctanx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}{ln}\mid\frac{{x}−\sqrt{\mathrm{2}}}{{x}+\sqrt{\mathrm{2}}}\mid+\frac{\mathrm{1}}{\mathrm{3}}{arctanx}\:+{C} \\ $$

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