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Question Number 86853 by M±th+et£s last updated on 01/Apr/20

∫((x^6 −x^3 +2)/(x^4 −x^2 −2))dx

$$\int\frac{{x}^{\mathrm{6}} −{x}^{\mathrm{3}} +\mathrm{2}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}{dx} \\ $$

Answered by MJS last updated on 01/Apr/20

((x^6 −x^3 +2)/(x^4 −x^2 −2))= =x^2 +1−((x^3 −3x^2 −4)/(x^4 −x^2 −2)) ((x^3 −3x^2 −4)/(x^4 −x^2 −2))= =−((x/(3(x^2 +1)))+(1/(3(x^2 +1)))+((2x)/(3(x^2 −2)))−((5(√2))/(6(x−(√2))))+((5(√2))/(6(x+(√2))))) ∫((x^6 −x^3 +2)/(x^4 −x^2 −2))dx= =∫x^2 dx+∫dx−(1/6)∫((2x)/(x^2 +1))dx−(1/3)∫(dx/(x^2 +1))−(1/3)∫((2x)/(x^2 −2))dx+((5(√2))/6)∫(dx/(x−(√2)))−((5(√2))/6)∫(dx/(x+(√2)))= =(1/3)x^3 +x−(1/6)ln (x^2 +1) −(1/3)arctan x −(1/3)ln (x^2 −2) +((5(√2))/6)ln (x−(√2)) −((5(√2))/6)ln (x+(√2)) = =((x^3 +3x)/3)−(1/6)ln ((x^2 +1)(x^2 −2)^2 ) +((5(√2))/6)ln ∣((x−(√2))/(x+(√2)))∣ −(1/3)arctan x +C

$$\frac{{x}^{\mathrm{6}} −{x}^{\mathrm{3}} +\mathrm{2}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}= \\ $$$$={x}^{\mathrm{2}} +\mathrm{1}−\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}= \\ $$$$=−\left(\frac{{x}}{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}{x}}{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}\right)}−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}\left({x}−\sqrt{\mathrm{2}}\right)}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}\left({x}+\sqrt{\mathrm{2}}\right)}\right) \\ $$$$\int\frac{{x}^{\mathrm{6}} −{x}^{\mathrm{3}} +\mathrm{2}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}{dx}= \\ $$$$=\int{x}^{\mathrm{2}} {dx}+\int{dx}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{2}}{dx}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}}\int\frac{{dx}}{{x}−\sqrt{\mathrm{2}}}−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}}\int\frac{{dx}}{{x}+\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +{x}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:{x}\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}\right)\:+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}}\mathrm{ln}\:\left({x}−\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}}\mathrm{ln}\:\left({x}+\sqrt{\mathrm{2}}\right)\:= \\ $$$$=\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left(\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \right)\:+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{6}}\mathrm{ln}\:\mid\frac{{x}−\sqrt{\mathrm{2}}}{{x}+\sqrt{\mathrm{2}}}\mid\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:{x}\:+{C} \\ $$

Commented by M±th+et£s last updated on 01/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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