Question Number 196965 by sonukgindia last updated on 05/Sep/23
Answered by qaz last updated on 05/Sep/23
$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}}\right)}\overset{{x}\rightarrow{x}^{\mathrm{4}} } {=}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}\right)} \\ $$$$\overset{{x}\rightarrow\frac{\mathrm{1}}{{x}}} {=}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{x}^{\mathrm{2}} {dx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}\right)}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{xdx}}{\mathrm{1}+{x}^{\mathrm{4}} }=\frac{\pi}{\mathrm{2}} \\ $$