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Question Number 66326 by mathmax by abdo last updated on 12/Aug/19

$$calculate{\int }_0^1\frac{x^4+1}{x^6+1}dx$$

Commented by Prithwish sen last updated on 13/Aug/19

$$\frac{{\mathrm{x}}^4+1}{{\mathrm{x}}^6+1}=\frac{\mathrm{A}}{{\mathrm{x}}^2+1}+\frac{\mathrm{B}}{{\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1}+\frac{\mathrm{C}}{{\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1}$$$${\mathrm{x}}^4+1={\mathrm{x}}^4(\mathrm{A}+\mathrm{B}+\mathrm{C})+{\mathrm{x}}^3(\sqrt{3}\mathrm{C}-\sqrt{3}\mathrm{B})+{\mathrm{x}}^2(2\mathrm{B}+2\mathrm{C}-\mathrm{A})+\mathrm{x}(\sqrt{3}\mathrm{C}-\sqrt{3}\mathrm{B})+(\mathrm{A}+\mathrm{B}+\mathrm{C})$$$$\Rightarrow \mathrm{A}+\mathrm{B}+\mathrm{C}=1$$$$\sqrt{3}\mathrm{C}-\sqrt{3}\mathrm{B}=0$$$$2\mathrm{B}+2\mathrm{C}-\mathrm{A}=0$$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=0$$$$\mathrm{A}=\frac{2}{3}\mathrm{B}=\frac{1}{6}\mathrm{C}=\frac{1}{6}$$$$\int \frac{2\mathrm{dx}}{3(1+{\mathrm{x}}^2)}+\frac{1}{6}\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{\sqrt{3}}{2})}^2+{\left(\frac{1}{2}\right)}^2}+\frac{1}{6}\int \frac{\mathrm{dx}}{{(\mathrm{x}-\frac{\sqrt{3}}{2})}^2+{\left(\frac{1}{2}\right)}^2}$$

Commented by mathmax by abdo last updated on 13/Aug/19

$$thankyousir.$$

Commented by Prithwish sen last updated on 13/Aug/19

$$\mathrm{welome}$$

Commented by mathmax by abdo last updated on 13/Aug/19

$${\int }_0^{+\mathrm{\infty }}\frac{x^4+1}{x^6+1}dx={\int }_0^1\frac{x^4+1}{x^6+1}dx+{\int }_1^{+\mathrm{\infty }}\frac{x^4+1}{x^6+1}dxbut$$$${\int }_1^{+\mathrm{\infty }}\frac{x^4+1}{x^6+1}dx{=}_{x=\frac{1}{t}}-{\int }_0^1\frac{\frac{1}{t^4}+1}{\frac{1}{t^6}+1}(-\frac{dt}{t^2})={\int }_0^1\frac{1+t^4}{1+t^6}dt\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{x^4+1}{x^6+1}dx=2{\int }_0^1\frac{x^4+1}{x^6+1}dx\Rightarrow {\int }_0^1\frac{x^4+1}{x^6+1}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^6}$$$$changementx=t^{\frac{1}{6}}give{\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{2}{3}}}{1+t}\frac{1}{6}{t}^{\frac{1}{6}-1}dt$$$$=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{2}{3}+\frac{1}{6}-1}}{1+t}dt=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{5}{6}-1}}{1+t}dt=\frac{1}{6}\frac{\pi }{sin\left(\frac{5\pi }{6}\right)}=\frac{\pi }{6}\times 2=\frac{\pi }{3}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^6}{=}_{x=t^{\frac{1}{6}}}\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{6}-1}}{1+t}dt=\frac{1}{6}\frac{\pi }{sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{6}\times 2=\frac{\pi }{3}\Rightarrow$$$${\int }_0^1\frac{x^4+1}{x^6+1}dx=\frac{\pi }{6}+\frac{\pi }{6}=\frac{\pi }{3}$$

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