Question Number 114671 by soumyasaha last updated on 20/Sep/20

   Integrate  ∫ ((x^4 +x^(−4) )/(x^6 +x^(−6) ))dx

Answered by Olaf last updated on 20/Sep/20

∫((x^(10) +x^2 )/(x^(12) +1))dx  x^(12) +1 = 0 ⇔ x^(12)  = −1 = e^(iπ)   x_k  = e^(i((π/(12))+((kπ)/6))) , k = 0,1,2...11  ∫Σ_(k=0) ^(11) (A_k /(x−x_k ))dx  A_k  = ((x_k ^6 (x_k ^4 +x_k ^(−4) ))/(Π_(j=0_(j≠k) ) ^(11) (x_j −x_k )))  x_k ^6 = e^(i((π/(12))+((kπ)/6))×6)  = e^(i((π/2)+kπ) = (−1)^k i)   x_k ^4 +x_k ^(−4)  = 2cos[((π/(12))+((kπ)/6))×4] = 2cos((π/3)+((2kπ)/3))  x_k ^4 +x_k ^(−4)  = 0 if k = 1,4,7,10  x_j −x_k  = e^(i((π/(12))+((jπ)/6))) −e^(i((π/(12))+((kπ)/6)))   x_j −x_k  = e^(i((π/(12))+(((j+k)π)/(12)))) [e^(i(((j−k)π)/(12))) −e^(−i(((j−k)π)/(12))) ]  x_j −x_k  = 2ie^(i((π/(12))+(((j+k)π)/(12)))) sin[(((j−k)π)/(12))]  A_k  = (((−1)^k i×2cos((π/3)+((2kπ)/3)))/(2ie^(i((π/(12))+(((j+k)π)/(12)))) sin[(((j−k)π)/(12))]))  A_k  = (((−1)^k cos((π/3)+((2kπ)/3)))/(sin[(((j−k)π)/(12))]))e^(−i((π/(12))+(((j+k)π)/(12))))   ∫((x^4 +x^(−4) )/(x^6 +x^(−6) ))dx = Σ_(k=0,2,3,5,6,8,9,11) A_k ln∣x−x_k ∣

Commented bysoumyasaha last updated on 20/Sep/20

Thanks Sir.