Question Number 80515 by M±th+et£s last updated on 03/Feb/20
$$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\phi} \right)^{\phi} } \\ $$
Commented by john santu last updated on 04/Feb/20
$${t}\:=\:\mathrm{1}+{x}^{\phi} \:\Rightarrow{x}=\left({t}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\phi}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\phi} }.\frac{\mathrm{1}}{\phi}\left({t}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\phi}−\mathrm{1}} {dt}\: \\ $$$$=\:\frac{\mathrm{1}}{\phi}\int_{\mathrm{0}} ^{\infty} {t}^{−\phi+\phi^{−\mathrm{1}} } \:\left(\mathrm{1}−{t}^{−\mathrm{1}} \right)^{\frac{\mathrm{1}}{\phi}−\mathrm{1}} {dt} \\ $$$$=\:\frac{\mathrm{1}}{\phi}.\frac{\Gamma\left(\phi−\frac{\mathrm{1}}{\phi}\right)\Gamma\left(\frac{\mathrm{1}}{\phi}\right)}{\Gamma\left(\phi\right)} \\ $$$$=\:\frac{\mathrm{1}}{\phi}.\frac{\Gamma\left(\phi^{−\mathrm{1}} \right)}{\phi^{−\mathrm{1}} \Gamma\left(\phi^{\left.−\mathrm{1}\right)} \right)}\:=\:\mathrm{1} \\ $$$${where}\:\phi=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}} \\ $$