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

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

Answered by Olaf last updated on 20/Sep/20

$$\int \frac{x^{10}+x^2}{x^{12}+1}dx$$$$x^{12}+1=0\iff x^{12}=-1=e^{i\pi }$$$$x_k=e^{i(\frac{\pi }{12}+\frac{k\pi }{6})},k=0,1,2...11$$$$\int \underset{k=0}{\sum ^{11}}\frac{{\mathrm{A}}_k}{x-x_k}dx$$$${\mathrm{A}}_k=\frac{x_k^6(x_k^4+x_k^{-4})}{\underset{\underset{j\ne k}{j=0}}{\prod ^{11}}(x_j-x_k)}$$$$x_k^6=e^{i(\frac{\pi }{12}+\frac{k\pi }{6})\times 6}=e^{i(\frac{\pi }{2}+k\pi )={(-1)}^{k}i}$$$$x_k^4+x_k^{-4}=2\cos \left[(\frac{\pi }{12}+\frac{k\pi }{6})\times 4\right]=2\cos (\frac{\pi }{3}+\frac{2k\pi }{3})$$$$x_k^4+x_k^{-4}=0\mathrm{if}k=1,4,7,10$$$$x_j-x_k=e^{i(\frac{\pi }{12}+\frac{j\pi }{6})}-e^{i(\frac{\pi }{12}+\frac{k\pi }{6})}$$$$x_j-x_k=e^{i(\frac{\pi }{12}+\frac{(j+k)\pi }{12})}[e^{i\frac{(j-k)\pi }{12}}-e^{-i\frac{(j-k)\pi }{12}}]$$$$x_j-x_k=2{ie}^{i(\frac{\pi }{12}+\frac{(j+k)\pi }{12})}\sin \left[\frac{(j-k)\pi }{12}\right]$$$${\mathrm{A}}_k=\frac{{(-1)}^{k}i\times 2\cos (\frac{\pi }{3}+\frac{2k\pi }{3})}{2{ie}^{i(\frac{\pi }{12}+\frac{(j+k)\pi }{12})}\sin \left[\frac{(j-k)\pi }{12}\right]}$$$${\mathrm{A}}_k=\frac{{(-1)}^k\cos (\frac{\pi }{3}+\frac{2k\pi }{3})}{\sin \left[\frac{(j-k)\pi }{12}\right]}{e}^{-i(\frac{\pi }{12}+\frac{(j+k)\pi }{12})}$$$$\int \frac{x^4+x^{-4}}{x^6+x^{-6}}dx=\sum _{k=0,2,3,5,6,8,9,11}{\mathrm{A}}_k\ln \mid x-x_k\mid$$$$$$

Commented by soumyasaha last updated on 20/Sep/20

$$\mathrm{Thanks}\mathrm{Sir}.$$

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