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Question Number 72062 by ozodbek last updated on 23/Oct/19

Commented by ozodbek last updated on 24/Oct/19

$solve please$
Commented by mathmax by abdo last updated on 24/Oct/19
$complex method z^5 +1 =0 ⇔z^5 =−1=e^((2k+1)π) so the roots are z_k =e^(i(((2k+1)π)/5)) with 0≤k≤4 ⇒(1/(x^5 +1)) =(1/(Π_(k=0) ^4 (x−z_k ))) =Σ_(k=0) ^4 (α_k /(x−z_k )) and α_k =(1/(5z_k ^4 )) =(z_k /(5z_k ^5 )) =−(z_k /5) ⇒ (1/(x^5 +1)) =−(1/5)Σ_(k=0) ^4 (z_k /(x−z_k )) =−(1/5){(z_o /(x−z_0 )) +(z_1 /(x−z_1 )) +(z_2 /(x−z_2 )) +(z_3 /(x−z_3 ))+(z_4 /(x−z_4 ))} ⇒ ∫ (dx/(x^5 +1)) =−(z_0 /5)ln(x−z_0 )−(z_1 /5)ln(x−z_1 )−(z_2 /5)ln(x−z_2 )−(z_3 /5)ln(x−z_2 ) −(z_4 /5)ln(x−z_2 )+c z_0 =e^((iπ)/5) , z_1 =e^(i((3π)/5)) ,z_2 =−1 , z_3 =e^((i7π)/5) ,z_4 =e^(i((9π)/5))$
$c o m p l e x m e t h o d z^{5} + 1 = 0 \Leftrightarrow z^{5} = - 1 = e^{(2 k + 1) π} s o t h e r o o t s a r e$ $z_{k} = e^{i \frac{(2 k + 1) π}{5}} w i t h 0 ⩽ k ⩽ 4 \Rightarrow \frac{1}{x^{5} + 1} = \frac{1}{\prod_{k = 0}^{4} (x - z_{k})}$ $= \sum_{k = 0}^{4} \frac{α_{k}}{x - z_{k}} a n d α_{k} = \frac{1}{5 z_{k}^{4}} = \frac{z_{k}}{5 z_{k}^{5}} = - \frac{z_{k}}{5} \Rightarrow$ $\frac{1}{x^{5} + 1} = - \frac{1}{5} \sum_{k = 0}^{4} \frac{z_{k}}{x - z_{k}} = - \frac{1}{5} {\frac{z_{o}}{x - z_{0}} + \frac{z_{1}}{x - z_{1}} + \frac{z_{2}}{x - z_{2}} + \frac{z_{3}}{x - z_{3}} + \frac{z_{4}}{x - z_{4}}} \Rightarrow$ $\int \frac{d x}{x^{5} + 1} = - \frac{z_{0}}{5} l n (x - z_{0}) - \frac{z_{1}}{5} l n (x - z_{1}) - \frac{z_{2}}{5} l n (x - z_{2}) - \frac{z_{3}}{5} l n (x - z_{2})$ $- \frac{z_{4}}{5} l n (x - z_{2}) + c$ $z_{0} = e^{\frac{i π}{5}}, z_{1} = e^{i \frac{3 π}{5}}, z_{2} = - 1, z_{3} = e^{\frac{i 7 π}{5}}, z_{4} = e^{i \frac{9 π}{5}}$
	Answered by ajfour last updated on 24/Oct/19

	$x^{5} + 1 = (x + 1) (x + a) (x + b) (x + c) (x + d)$ $I = k \ln ∣ x + 1 ∣ + Σ A \ln ∣ x + a ∣$ $k = \frac{1}{(a - 1) (b - 1) (c - 1) (d - 1)}$ $A = \frac{1}{(1 - a) (b - a) (c - a) (d - a)}$ $a = e^{2 i π / 5}, b = e^{4 i π / 5}, c = e^{6 i π / 5}, d = e^{8 i π / 5} .$
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