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.