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Question Number 65332 by aliesam last updated on 28/Jul/19

Commented by mathmax by abdo last updated on 29/Jul/19
$let suppose n inter let decompose the fraction F(x) =(x/(x^n +1)) z^n +1 =0 ⇒z^n =−1 ⇒r^n e^(inθ) =e^((2k+1)π) (z=re^(iθ) ) ⇒r=1 andθ θ =(((2k+1)π)/n) so the roots are Z_k =e^(i(((2k+1)π)/n)) k∈[[0,n−1]] F(x) =Σ_(i=0) ^(n−1) (λ_i /(x−Z_i )) with λ_i =(Z_i /(nZ_i ^(n−1) )) =(Z_i ^2 /(−n)) ⇒ F(x) =−(1/n)Σ_(i=0) ^(n−1) (Z_i ^2 /(x−Z_i )) ⇒ ∫ F(x)dx =−(1/n)Σ_(i=0) ^(n−1) Z_i ^2 ∫ (dx/(x−Z_i )) rest to determine ∫ (dx/(x−Z_i )) be continued...$
$l e t s u p p o s e n i n t e r l e t d e c o m p o s e t h e f r a c t i o n$ $F (x) = \frac{x}{x^{n} + 1}$ $z^{n} + 1 = 0 \Rightarrow z^{n} = - 1 \Rightarrow r^{n} e^{i n θ} = e^{(2 k + 1) π} (z = {r e}^{i θ}) \Rightarrow r = 1 a n d θ$ $θ = \frac{(2 k + 1) π}{n} s o t h e r o o t s a r e Z_{k} = e^{i \frac{(2 k + 1) π}{n}} k \in [[0, n - 1]]$ $F (x) = \sum_{i = 0}^{n - 1} \frac{λ_{i}}{x - Z_{i}} w i t h λ_{i} = \frac{Z_{i}}{{n Z}_{i}^{n - 1}} = \frac{Z_{i}^{2}}{- n} \Rightarrow$ $F (x) = - \frac{1}{n} \sum_{i = 0}^{n - 1} \frac{Z_{i}^{2}}{x - Z_{i}} \Rightarrow \int F (x) d x = - \frac{1}{n} \sum_{i = 0}^{n - 1} Z_{i}^{2} \int \frac{d x}{x - Z_{i}}$ $r e s t t o d e t e r m i n e \int \frac{d x}{x - Z_{i}} b e c o n t i n u e d . . .$
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