Integrate-x-4-x-4-x-6-x-6-dx-

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

Integrate \int \frac{x^{4} + x^{- 4}}{x^{6} + x^{- 6}} dx

Answered by Olaf last updated on 20/Sep/20

\int \frac{x^{10} + x^{2}}{x^{12} + 1} d x

x^{12} + 1 = 0 \Leftrightarrow x^{12} = - 1 = e^{i π}

x_{k} = e^{i (\frac{π}{12} + \frac{k π}{6})}, k = 0, 1, 2 \dots 11

\int \underset{k = 0}{\sum^{11}} \frac{A_{k}}{x - x_{k}} d x

A_{k} = \frac{x_{k}^{6} (x_{k}^{4} + x_{k}^{- 4})}{\underset{\underset{j \neq k}{j = 0}}{\prod^{11}} (x_{j} - x_{k})}

x_{k}^{6} = e^{i (\frac{π}{12} + \frac{k π}{6}) \times 6} = e^{i (\frac{π}{2} + k π) = {(- 1)}^{k} i}

x_{k}^{4} + x_{k}^{- 4} = 2 \cos [(\frac{π}{12} + \frac{k π}{6}) \times 4] = 2 \cos (\frac{π}{3} + \frac{2 k π}{3})

x_{k}^{4} + x_{k}^{- 4} = 0 if k = 1, 4, 7, 10

x_{j} - x_{k} = e^{i (\frac{π}{12} + \frac{j π}{6})} - e^{i (\frac{π}{12} + \frac{k π}{6})}

x_{j} - x_{k} = e^{i (\frac{π}{12} + \frac{(j + k) π}{12})} [e^{i \frac{(j - k) π}{12}} - e^{- i \frac{(j - k) π}{12}}]

x_{j} - x_{k} = 2 {i e}^{i (\frac{π}{12} + \frac{(j + k) π}{12})} \sin [\frac{(j - k) π}{12}]

A_{k} = \frac{{(- 1)}^{k} i \times 2 \cos (\frac{π}{3} + \frac{2 k π}{3})}{2 {i e}^{i (\frac{π}{12} + \frac{(j + k) π}{12})} \sin [\frac{(j - k) π}{12}]}

A_{k} = \frac{{(- 1)}^{k} \cos (\frac{π}{3} + \frac{2 k π}{3})}{\sin [\frac{(j - k) π}{12}]} e^{- i (\frac{π}{12} + \frac{(j + k) π}{12})}

\int \frac{x^{4} + x^{- 4}}{x^{6} + x^{- 6}} d x = \sum_{k = 0, 2, 3, 5, 6, 8, 9, 11} A_{k} \ln ∣ x - x_{k} ∣

Commented by soumyasaha last updated on 20/Sep/20

Thanks Sir .