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Question-72062

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Question Number 72062 by ozodbek last updated on 23/Oct/19
Commented by ozodbek last updated on 24/Oct/19
solve please
$$\mathrm{solve}\:\mathrm{please}\: \\ $$
Commented by mathmax by abdo last updated on 24/Oct/19
complex method z^5 +1 =0 ⇔z^5 =−1=e^((2k+1)π) so the roots are z_k =e^(i(((2k+1)π)/5)) with 0≤k≤4 ⇒(1/(x^5 +1)) =(1/(Π_(k=0) ^4 (x−z_k ))) =Σ_(k=0) ^4 (α_k /(x−z_k )) and α_k =(1/(5z_k ^4 )) =(z_k /(5z_k ^5 )) =−(z_k /5) ⇒ (1/(x^5 +1)) =−(1/5)Σ_(k=0) ^4 (z_k /(x−z_k )) =−(1/5){(z_o /(x−z_0 )) +(z_1 /(x−z_1 )) +(z_2 /(x−z_2 )) +(z_3 /(x−z_3 ))+(z_4 /(x−z_4 ))} ⇒ ∫ (dx/(x^5 +1)) =−(z_0 /5)ln(x−z_0 )−(z_1 /5)ln(x−z_1 )−(z_2 /5)ln(x−z_2 )−(z_3 /5)ln(x−z_2 ) −(z_4 /5)ln(x−z_2 )+c z_0 =e^((iπ)/5) , z_1 =e^(i((3π)/5)) ,z_2 =−1 , z_3 =e^((i7π)/5) ,z_4 =e^(i((9π)/5))
$${complex}\:{method}\:\:\:\:{z}^{\mathrm{5}} +\mathrm{1}\:=\mathrm{0}\:\Leftrightarrow{z}^{\mathrm{5}} =−\mathrm{1}={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{roots}\:{are} \\ $$$${z}_{{k}} ={e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}}} \:\:{with}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{4}\:\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}\:=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{\mathrm{4}} \left({x}−{z}_{{k}} \right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{\alpha_{{k}} }{{x}−{z}_{{k}} }\:\:{and}\:\alpha_{{k}} =\frac{\mathrm{1}}{\mathrm{5}{z}_{{k}} ^{\mathrm{4}} }\:=\frac{{z}_{{k}} }{\mathrm{5}{z}_{{k}} ^{\mathrm{5}} }\:=−\frac{{z}_{{k}} }{\mathrm{5}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{1}}\:=−\frac{\mathrm{1}}{\mathrm{5}}\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:=−\frac{\mathrm{1}}{\mathrm{5}}\left\{\frac{{z}_{{o}} }{{x}−{z}_{\mathrm{0}} }\:+\frac{{z}_{\mathrm{1}} }{{x}−{z}_{\mathrm{1}} }\:+\frac{{z}_{\mathrm{2}} }{{x}−{z}_{\mathrm{2}} }\:+\frac{{z}_{\mathrm{3}} }{{x}−{z}_{\mathrm{3}} }+\frac{{z}_{\mathrm{4}} }{{x}−{z}_{\mathrm{4}} }\right\}\:\Rightarrow \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{5}} +\mathrm{1}}\:=−\frac{{z}_{\mathrm{0}} }{\mathrm{5}}{ln}\left({x}−{z}_{\mathrm{0}} \right)−\frac{{z}_{\mathrm{1}} }{\mathrm{5}}{ln}\left({x}−{z}_{\mathrm{1}} \right)−\frac{{z}_{\mathrm{2}} }{\mathrm{5}}{ln}\left({x}−{z}_{\mathrm{2}} \right)−\frac{{z}_{\mathrm{3}} }{\mathrm{5}}{ln}\left({x}−{z}_{\mathrm{2}} \right) \\ $$$$−\frac{{z}_{\mathrm{4}} }{\mathrm{5}}{ln}\left({x}−{z}_{\mathrm{2}} \right)+{c} \\ $$$${z}_{\mathrm{0}} ={e}^{\frac{{i}\pi}{\mathrm{5}}} \:\:,\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{5}}} \:\:,{z}_{\mathrm{2}} =−\mathrm{1}\:,\:{z}_{\mathrm{3}} ={e}^{\frac{{i}\mathrm{7}\pi}{\mathrm{5}}} \:\:,{z}_{\mathrm{4}} ={e}^{{i}\frac{\mathrm{9}\pi}{\mathrm{5}}} \\ $$
Answered by ajfour last updated on 24/Oct/19
x^5 +1=(x+1)(x+a)(x+b)(x+c)(x+d) I=kln ∣x+1∣+ΣAln ∣x+a∣ k=(1/((a−1)(b−1)(c−1)(d−1))) A=(1/((1−a)(b−a)(c−a)(d−a))) a=e^(2iπ/5) , b=e^(4iπ/5) , c=e^(6iπ/5) , d=e^(8iπ/5) .
$${x}^{\mathrm{5}} +\mathrm{1}=\left({x}+\mathrm{1}\right)\left({x}+{a}\right)\left({x}+{b}\right)\left({x}+{c}\right)\left({x}+{d}\right) \\ $$$${I}={k}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid+\Sigma{A}\mathrm{ln}\:\mid{x}+{a}\mid \\ $$$$\:\:\:{k}=\frac{\mathrm{1}}{\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({d}−\mathrm{1}\right)} \\ $$$$\:\:{A}=\frac{\mathrm{1}}{\left(\mathrm{1}−{a}\right)\left({b}−{a}\right)\left({c}−{a}\right)\left({d}−{a}\right)} \\ $$$$\:\:{a}={e}^{\mathrm{2}{i}\pi/\mathrm{5}} \:,\:{b}={e}^{\mathrm{4}{i}\pi/\mathrm{5}} ,\:{c}={e}^{\mathrm{6}{i}\pi/\mathrm{5}} ,\:{d}={e}^{\mathrm{8}{i}\pi/\mathrm{5}} . \\ $$

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