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Question Number 67528 by mathmax by abdo last updated on 28/Aug/19

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1+x^3}{1+x^6}dx$$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$$letnameditK$$$$whenchangingx=-u$$$$K={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-u^3}{1+u^6}du$$$$So2K={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{2du}{1+u^6}=4{\int }_0^{\mathrm{\infty }}\frac{du}{1+u^6}$$$$changeu=v^{\frac{1}{6}}\Rightarrow du=\frac{v^{\frac{1}{6}-1}}{6}dv$$$$K=\frac{2}{6}{\int }_0^{\mathrm{\infty }}\frac{v^{\frac{1}{6}-1}}{1+v}dv=\frac{\pi }{3sin\left(\frac{\pi }{6}\right)}=\frac{2\pi }{3}$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$thankyousir.$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1+x^3}{1+x^6}dx\Rightarrow I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{1+x^6}+{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^3}{1+x^6}dx$$$$but{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^3}{1+x^6}dx=0\left(thefunctionisodd\right)\Rightarrow I=2{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^6}$$$$cha7gementx=t^{\frac{1}{6}}giveI=2{\int }_0^{\mathrm{\infty }}\frac{1}{1+t}\times \frac{1}{6}{t}^{\frac{1}{6}-1}dt$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{6}-1}}{1+t}dt=\frac{1}{3}\frac{\pi }{sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{3\times \frac{1}{2}}=\frac{2\pi }{3}.$$

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