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Question Number 40131 by maxmathsup by imad last updated on 16/Jul/18

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

Answered by maxmathsup by imad last updated on 24/Jul/18

$$wehave{\int }_0^1\frac{dx}{1+x^6}{=}_{x=\frac{1}{t}}{\int }_1^{+\mathrm{\infty }}\frac{1}{1+\frac{1}{t^6}}\frac{dt}{t^2}={\int }_1^{+\mathrm{\infty }}\frac{dt}{t^2+\frac{1}{t^4}}$$$$={\int }_1^{+\mathrm{\infty }}\frac{t^4}{1+t^6}dt={\int }_0^{\mathrm{\infty }}\frac{t^4}{1+t^6}dt-{\int }_0^1\frac{t^4}{1+t^6}dt\Rightarrow$$$${\int }_0^1\frac{1+x^4}{1+x^6}dx={\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx{=}_{x=u^{\frac{1}{6}}}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{4}{6}}}{1+u}\frac{1}{6}{u}^{\frac{1}{6}-1}du$$$$=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{2}{3}+\frac{1}{6}-1}}{1+u}du=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{5}{6}-1}}{1+u}du$$$$=\frac{1}{6}\frac{\pi }{sin\left(\frac{5\pi }{6}\right)}=\frac{\pi }{6sin(\pi -\frac{\pi }{6})}=\frac{\pi }{6sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{6.\frac{1}{2}}\Rightarrow I=\frac{\pi }{3}.$$

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