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Question-76910

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Question Number 76910 by aliesam last updated on 31/Dec/19
Commented by MJS last updated on 31/Dec/19
x^6 +1=(x^2 +1)(x^2 +(√3)x+1)(x^2 −(√3)x+1)
$${x}^{\mathrm{6}} +\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right) \\ $$
Answered by MJS last updated on 01/Jan/20
∫(dx/(x^6 +1))= =(1/3)∫(dx/(x^2 +1))−(1/6)∫(((√3)x−2)/(x^2 −(√3)x+1))dx+(1/6)∫(((√3)x+2)/(x^2 +(√3)x+1))dx I_1 =(1/3)∫(dx/(x^2 +1))=(1/3)arctan x I_2 =−(1/6)∫(((√3)x−2)/(x^2 −(√3)x+1))dx= =−((√3)/(12))∫((2x−(√3))/(x^2 −(√3)x+1))dx+(1/(12))∫(dx/(x^2 −(√3)x+1))= =−((√3)/(12))ln (x^2 −(√3)x+1) +(1/6)arctan (2x−(√3)) I_3 =(1/6)∫(((√3)x+2)/(x^2 +(√3)x+1))dx= =((√3)/(12))∫((2x+(√3))/(x^2 +(√3)x+1))dx+(1/(12))∫(dx/(x^2 +(√3)x+1))= =((√3)/(12))ln (x^2 +(√3)x+1) +(1/6)arctan (2x+(√3)) I=I_1 +I_2 +I_3 +C
$$\int\frac{{dx}}{{x}^{\mathrm{6}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:{x} \\ $$$$\:\:\:\:\:{I}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\int\frac{\mathrm{2}{x}−\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:{I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\sqrt{\mathrm{3}}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\int\frac{\mathrm{2}{x}+\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right) \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} +{I}_{\mathrm{3}} +{C} \\ $$

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