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Question Number 42695 by prof Abdo imad last updated on 31/Aug/18

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^4}{x^8+16}dx$$

Commented by maxmathsup by imad last updated on 01/Sep/18

$$letA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^4}{x^8+16}dx\Rightarrow A=2{\int }_0^{\mathrm{\infty }}\frac{x^4}{x^8+{\left(\sqrt{2}\right)}^8}changementx=t\sqrt{2}give$$$$A=2{\int }_0^{\mathrm{\infty }}\frac{4t^4}{16t^8+16}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^4}{1+t^8}dt$$$${=}_{t=u^{\frac{1}{8}}}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}}}{1+u}\frac{1}{8}{u}^{\frac{1}{8}-1}du=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}+\frac{1}{8}-1}}{1+u}du$$$$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{5}{8}-1}}{1+u}du=\frac{1}{16}\frac{\pi }{sin\left(\frac{5\pi }{8}\right)}butsin\left(\frac{5\pi }{8}\right)=sin(\frac{\pi }{2}+\frac{\pi }{8})=cos\left(\frac{\pi }{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}$$$$\Rightarrow A=\frac{\pi }{16}\frac{\sqrt{2+\sqrt{2}}}{2}\Rightarrow A=\frac{\pi \sqrt{2+\sqrt{2}}}{32}.$$

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