dx-1-x-12-

Question Number 90417 by jagoll last updated on 23/Apr/20

$$\int\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{12}} } \\ $$

Commented by MJS last updated on 23/Apr/20

I′m afraid you will have to find 6 square factors of x^(12) +1 and then decompose...

$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\mathrm{6}\:\mathrm{square} \\ $$$$\mathrm{factors}\:\mathrm{of}\:{x}^{\mathrm{12}} +\mathrm{1}\:\mathrm{and}\:\mathrm{then}\:\mathrm{decompose}… \\ $$

Commented by jagoll last updated on 23/Apr/20

$$\mathrm{hahaha}…\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{scary}\:\mathrm{question}\: \\ $$$$\mathrm{sir}\: \\ $$

Commented by MJS last updated on 23/Apr/20

the real square factors of x^(12) +1 are x^2 −(√2)x+1 x^2 +(√2)x+1 x^2 −((((√6)+(√2))/2))x+1 x^2 +((((√6)+(√2))/2))x+1 x^2 −((((√6)−(√2))/2))x+1 x^2 +((((√6)−(√2))/2))x+1 ...I don′t want to decompose, sorry

$$\mathrm{the}\:\mathrm{real}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{of}\:{x}^{\mathrm{12}} +\mathrm{1}\:\mathrm{are} \\ $$$${x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right){x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right){x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right){x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right){x}+\mathrm{1} \\ $$$$…\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{decompose},\:\mathrm{sorry} \\ $$

Commented by jagoll last updated on 23/Apr/20

$$\mathrm{Ostrogradski}\:\mathrm{sir}? \\ $$

Commented by MJS last updated on 23/Apr/20

no. x^(12) +1=(x^4 +1)(x^8 −x^4 +1) x^4 +1=Π((√2)/2)(±1∓i) [all combinations of +/−] x^8 −x^4 +1=(x^4 )^2 −(x^4 )+1=Π(x^4 −(1/2)±((√3)/2)i) and then solve x^4 =(1/2)±((√3)/2)

$$\mathrm{no}. \\ $$$${x}^{\mathrm{12}} +\mathrm{1}=\left({x}^{\mathrm{4}} +\mathrm{1}\right)\left({x}^{\mathrm{8}} −{x}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} +\mathrm{1}=\Pi\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\pm\mathrm{1}\mp\mathrm{i}\right)\:\left[\mathrm{all}\:\mathrm{combinations}\:\mathrm{of}\:+/−\right] \\ $$$${x}^{\mathrm{8}} −{x}^{\mathrm{4}} +\mathrm{1}=\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} −\left({x}^{\mathrm{4}} \right)+\mathrm{1}=\Pi\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{solve}\:{x}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Commented by jagoll last updated on 23/Apr/20

oo (x^4 )^3 +1 = (x^4 +1)((x^4 )^2 −x^4 .1+1). thank you sir

$$\mathrm{oo}\:\left(\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{3}} \:+\mathrm{1}\:=\:\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\left(\left(\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{4}} .\mathrm{1}+\mathrm{1}\right).\: \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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