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Question Number 132000 by bramlexs22 last updated on 10/Feb/21

$$\mathrm{super}\mathrm{nice}$$$${\int }_0^{\mathrm{\infty }}\left(\frac{1}{{({\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3})}^2}\right)\mathrm{dx}$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\mathrm{E}=\int \frac{{\mathrm{x}}^6}{{({\mathrm{x}}^6+1)}^2}\mathrm{dx};\mathrm{apply}\mathrm{integration}\mathrm{by}\mathrm{parts}$$$$\{\begin{array}{l}\mathrm{u}=\mathrm{x}\Rightarrow \mathrm{du}=\mathrm{dx}\\ \mathrm{dv}=\frac{{\mathrm{x}}^5}{{({\mathrm{x}}^6+1)}^2}\Rightarrow \mathrm{v}=-\frac{1}{6}.\frac{1}{{\mathrm{x}}^6+1}\end{array}$$$${\mathrm{E}=-\frac{\mathrm{x}}{6({\mathrm{x}}^6+1)}]}_0^{\mathrm{\infty }}+\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^6+1}$$$$\mathrm{E}=0+\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^6+1};\mathrm{replacing}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}$$$$\mathrm{E}=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^4}{{\mathrm{x}}^6+1}\mathrm{dx};\mathrm{adding}\mathrm{we}\mathrm{get}$$$$2\mathrm{E}=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^4+1}{{\mathrm{x}}^6+1}\mathrm{dx}$$$$\mathrm{E}=\frac{1}{12}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+{\mathrm{x}}^{-2}}{{\mathrm{x}}^3+{\mathrm{x}}^{-3}}\left(\frac{\mathrm{dx}}{\mathrm{x}}\right)=\frac{1}{12}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+{\mathrm{x}}^{-2}}{{\mathrm{x}}^3+{\mathrm{x}}^{-3}}\frac{\mathrm{d}(\mathrm{x}-{\mathrm{x}}^{-1})}{\mathrm{x}+{\mathrm{x}}^{-1}}$$$$\mathrm{E}=\frac{1}{12}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+{\mathrm{x}}^{-2}}{{\mathrm{x}}^2-1+{\mathrm{x}}^{-2}}\frac{\mathrm{d}(\mathrm{x}-{\mathrm{x}}^{-1})}{{(\mathrm{x}+{\mathrm{x}}^{-1})}^2}$$$$\mathrm{letting}\mathrm{r}=\mathrm{x}-{\mathrm{x}}^{-1}$$$$\mathrm{E}=\frac{1}{12}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{r}}^2+2}{({\mathrm{r}}^2+1)({\mathrm{r}}^2+4)}\mathrm{dr}$$$$\mathrm{E}=\frac{1}{36}{\underset{-\mathrm{\infty }}{\int }}^{\mathrm{\infty }}(\frac{1}{{\mathrm{r}}^2+1}+\frac{2}{{\mathrm{r}}^2+4})\mathrm{dr}$$$$\mathrm{E}=\frac{1}{36}{(\arctan \mathrm{r}+\arctan \left(\frac{\mathrm{r}}{2}\right)]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{\pi }{18}$$$$$$

Answered by Ar Brandon last updated on 10/Feb/21

$$\mathcal{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3})}^2}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^6}{{({\mathrm{x}}^6+1)}^2}\mathrm{dx}$$$${\mathrm{x}}^6=\mathrm{u}\Rightarrow {6\mathrm{x}}^5\mathrm{dx}=\mathrm{du}\Rightarrow {6\mathrm{u}}^{\frac{5}{6}}\mathrm{dx}=\mathrm{du}$$$$\mathcal{I}=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{1}{6}}}{{(\mathrm{u}+1)}^2}\mathrm{du}=\frac{1}{6}\beta (\frac{7}{6},\frac{5}{6})$$$$=\frac{1}{6}\cdot \frac{1}{6}\cdot \frac{\pi }{\sin \left(\frac{\pi }{6}\right)}=\frac{\pi }{18}$$

Commented by bramlexs22 last updated on 10/Feb/21

$$\mathrm{with}\mathrm{betha}\mathrm{function}$$

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