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Question Number 86484 by Ar Brandon last updated on 28/Mar/20
∫(x^6 /(1+x^(12) ))dx
$$\int\frac{{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{12}} }{dx} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
Hi mathmax, I don't understand the second line of your solution. How did you do that ?
Commented by redmiiuser last updated on 29/Mar/20
sir can you pls check the answer
$${sir}\:{can}\:{you}\:{pls}\:{check} \\ $$$${the}\:{answer} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
How did you arrive there please ? ��
Commented by mathmax by abdo last updated on 29/Mar/20
z^(12) +1 =0 ⇒z^(12) =−1 =e^(i(2k+1)π) ⇒z_k =e^((i(2k+1)π)/(12)) k∈[[0,11] ⇒(x^6 /(x^(12) +1)) =(x^6 /(Π_(k=0) ^(11) (x−z_k ))) =Σ_(k=0) ^(11) (a_k /(x−z_k )) a_k =(z_k ^6 /(12z_k ^(11) )) =(z_k ^7 /(−12)) ⇒(x^6 /(x^(12) +1)) =−(1/(12))Σ_(k=0) ^(11) (z_k ^7 /(x−z_k )) ⇒ ∫ (x^6 /(x^(12) +1))dx =−(1/(12))Σ_(k=0) ^(11) z_k ^7 ln(x−z_k ) +C
$${z}^{\mathrm{12}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{z}^{\mathrm{12}} =−\mathrm{1}\:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow{z}_{{k}} ={e}^{\frac{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{12}}} \:\:{k}\in\left[\left[\mathrm{0},\mathrm{11}\right]\right. \\ $$$$\Rightarrow\frac{{x}^{\mathrm{6}} }{{x}^{\mathrm{12}} \:+\mathrm{1}}\:=\frac{{x}^{\mathrm{6}} }{\prod_{{k}=\mathrm{0}} ^{\mathrm{11}} \left({x}−{z}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{11}} \:\frac{{a}_{{k}} }{{x}−{z}_{{k}} } \\ $$$${a}_{{k}} =\frac{{z}_{{k}} ^{\mathrm{6}} }{\mathrm{12}{z}_{{k}} ^{\mathrm{11}} }\:=\frac{{z}_{{k}} ^{\mathrm{7}} }{−\mathrm{12}}\:\Rightarrow\frac{{x}^{\mathrm{6}} }{{x}^{\mathrm{12}} \:+\mathrm{1}}\:=−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{k}=\mathrm{0}} ^{\mathrm{11}} \:\frac{{z}_{{k}} ^{\mathrm{7}} }{{x}−{z}_{{k}} }\:\Rightarrow \\ $$$$\int\:\:\frac{{x}^{\mathrm{6}} }{{x}^{\mathrm{12}} +\mathrm{1}}{dx}\:=−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{k}=\mathrm{0}} ^{\mathrm{11}} \:{z}_{{k}} ^{\mathrm{7}} {ln}\left({x}−{z}_{{k}} \right)\:+{C} \\ $$
Commented by mathmax by abdo last updated on 29/Mar/20
its a like theorem if F =((p(x))/(Q(x))) with degp<degQ and Q without real roots (⇒Q(x) =λΠ_i (x−z_i )) so F(x)=Σ_i (a_i /(x−z_i )) and a_i =((p(z_i ))/(Q^′ (z_i )))
$${its}\:{a}\:{like}\:{theorem}\:{if}\:{F}\:=\frac{{p}\left({x}\right)}{{Q}\left({x}\right)}\:\:{with}\:{degp}<{degQ}\:{and} \\ $$$${Q}\:{without}\:{real}\:{roots}\:\left(\Rightarrow{Q}\left({x}\right)\:=\lambda\prod_{{i}} \left({x}−{z}_{{i}} \right)\right)\:{so} \\ $$$${F}\left({x}\right)=\sum_{{i}} \:\:\frac{{a}_{{i}} }{{x}−{z}_{{i}} }\:\:{and}\:{a}_{{i}} =\frac{{p}\left({z}_{{i}} \right)}{{Q}^{'} \left({z}_{{i}} \right)} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
Oh ! Thanks ��
Commented by Ar Brandon last updated on 29/Mar/20
what about a_k =(z_k ^6 /(12z_k ^(11) ))=(z_k ^7 /(−12)) ?
$$\mathrm{what}\:\mathrm{about}\:\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{6}} }{\mathrm{12z}_{\mathrm{k}} ^{\mathrm{11}} }=\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{7}} }{−\mathrm{12}}\:\:? \\ $$
Commented by mathmax by abdo last updated on 30/Mar/20
because z_k ^(12) =−1
$${because}\:{z}_{{k}} ^{\mathrm{12}} =−\mathrm{1} \\ $$
Commented by floor(10²Eta[1]) last updated on 09/Jul/20
hey mathmax where i can see the demonstration of this theorem?
$$\mathrm{hey}\:\mathrm{mathmax}\:\mathrm{where}\:\mathrm{i}\:\mathrm{can}\:\mathrm{see}\:\mathrm{the}\: \\ $$$$\mathrm{demonstration}\:\mathrm{of}\:\mathrm{this}\:\mathrm{theorem}? \\ $$
Answered by redmiiuser last updated on 29/Mar/20
(1+x^(12) )^(−1) =1+(−1)x^(12) +(((−1)(−1−1)x^(24) )/(2!))+(((−1)(−1−1)(−1−2)x^(36) )/(3!))+(((−1)(−1−1)(−1−2)(−1−3)x^(48) )/(4!))+....∞ =Σ_(n=0) ^∞ (((−1)^n .n!.(x^(12) )^n )/(n!)) =Σ_(n=0) ^∞ (−1)^n .(x^(12) )^n ∴x^6 .(1+x^(12) )^(−1) =x^6 .Σ_(n=0) ^∞ (−1)^n .(x^(12n) ) =Σ_(n=0) ^∞ (−1)^n .(x^(12n+6) ) ∴∫Σ_(n=0) ^∞ (−1)^n .(x^(12n+6) ).dx =Σ_(n=0) ^∞ (((−1)^n .(x^(12n+7) ))/(12n+7))
$$\left(\mathrm{1}+{x}^{\mathrm{12}} \right)^{−\mathrm{1}} \\ $$$$=\mathrm{1}+\left(−\mathrm{1}\right){x}^{\mathrm{12}} +\frac{\left(−\mathrm{1}\right)\left(−\mathrm{1}−\mathrm{1}\right){x}^{\mathrm{24}} }{\mathrm{2}!}+\frac{\left(−\mathrm{1}\right)\left(−\mathrm{1}−\mathrm{1}\right)\left(−\mathrm{1}−\mathrm{2}\right){x}^{\mathrm{36}} }{\mathrm{3}!}+\frac{\left(−\mathrm{1}\right)\left(−\mathrm{1}−\mathrm{1}\right)\left(−\mathrm{1}−\mathrm{2}\right)\left(−\mathrm{1}−\mathrm{3}\right){x}^{\mathrm{48}} }{\mathrm{4}!}+….\infty \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{n}!.\left({x}^{\mathrm{12}} \right)^{{n}} }{{n}!} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .\left({x}^{\mathrm{12}} \right)^{{n}} \\ $$$$\therefore{x}^{\mathrm{6}} .\left(\mathrm{1}+{x}^{\mathrm{12}} \right)^{−\mathrm{1}} \\ $$$$={x}^{\mathrm{6}} .\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .\left({x}^{\mathrm{12}{n}} \right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .\left({x}^{\mathrm{12}{n}+\mathrm{6}} \right) \\ $$$$\therefore\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .\left({x}^{\mathrm{12}{n}+\mathrm{6}} \right).{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left({x}^{\mathrm{12}{n}+\mathrm{7}} \right)}{\mathrm{12}{n}+\mathrm{7}} \\ $$
Commented by redmiiuser last updated on 29/Mar/20
welcome sir
$${welcome}\:{sir} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
Wow ! great idea. Thanks ! ��
Commented by redmiiuser last updated on 29/Mar/20
pls check the answer
$${pls}\:{check}\:{the}\:{answer} \\ $$
Answered by MJS last updated on 29/Mar/20
∫(x^6 /(x^(12) +1))dx= =(((√6)−(√2))/(24))∫(x/(x^2 −(((√6)+(√2))/2)x+1))dx− −(((√6)−(√2))/(24))∫(x/(x^2 +(((√6)+(√2))/2)x+1))dx+ +(((√6)+(√2))/(24))∫(x/(x^2 −(((√6)−(√2))/2)x+1))dx− −(((√6)+(√2))/(24))∫(x/(x^2 +(((√6)−(√2))/2)x+1))dx− −((√2)/(12))∫(x/(x^2 −(√2)x+1))dx+ +((√2)/(12))∫(x/(x^2 +(√2)x+1))dx now use formula
$$\int\frac{{x}^{\mathrm{6}} }{{x}^{\mathrm{12}} +\mathrm{1}}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{24}}\int\frac{{x}}{{x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}{x}+\mathrm{1}}{dx}− \\ $$$$−\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{24}}\int\frac{{x}}{{x}^{\mathrm{2}} +\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}}{x}+\mathrm{1}}{dx}+ \\ $$$$+\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{24}}\int\frac{{x}}{{x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}{x}+\mathrm{1}}{dx}− \\ $$$$−\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{24}}\int\frac{{x}}{{x}^{\mathrm{2}} +\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}{x}+\mathrm{1}}{dx}− \\ $$$$−\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\int\frac{{x}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+ \\ $$$$+\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\int\frac{{x}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{formula} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
I also used a method similar to yours and it was so bulky I made use of the fact that cos ((𝛑/(12)))=(((√6)+(√2))/4) and sin((𝛑/(12)))=(((√6)−(√2))/4) Was that the same idea which you used ?
$${I}\:{also}\:{used}\:{a}\:{method}\:{similar}\:{to}\:{yours}\:{and}\:{it}\:{was}\:{so}\:{bulky} \\ $$$${I}\:{made}\:{use}\:{of}\:{the}\:{fact}\:{that}\:\boldsymbol{{cos}}\:\left(\frac{\boldsymbol{\pi}}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${Was}\:{that}\:{the}\:{same}\:{idea}\:{which}\:{you}\:\:{used}\:? \\ $$
Commented by MJS last updated on 29/Mar/20
yes
$$\mathrm{yes} \\ $$

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