Question Number 161999 by mahdipoor last updated on 25/Dec/21
$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}} \\ $$
Answered by aleks041103 last updated on 25/Dec/21
$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}}=\int\frac{{dx}}{\:\sqrt{{x}}}\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}= \\ $$$$=\mathrm{2}\int\frac{{d}\left(\sqrt{{x}}\right)}{\:\sqrt{\left(\sqrt{{x}}\right)^{\mathrm{4}} −\mathrm{4}}}=\mathrm{2}\int\frac{{d}\left(\sqrt{{x}}\right)}{\:\mathrm{2}\sqrt{\left(\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{4}} −\mathrm{1}}}= \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{du}}{\:\sqrt{{u}^{\mathrm{4}} −\mathrm{1}}} \\ $$
Commented by aleks041103 last updated on 25/Dec/21
$${this}\:{integral}\:{cannot}\:{be}\:{solved}\:{using}\:{elementary}\:{functions} \\ $$
Commented by Ar Brandon last updated on 25/Dec/21
$$=−\sqrt{\mathrm{2}}{i}\int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }}=−\sqrt{\mathrm{2}}{i}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{u}^{\mathrm{4}{n}} {du} \\ $$$$=−\sqrt{\mathrm{2}}{i}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{4}{n}+\mathrm{1}\right)}{u}^{\mathrm{4}{n}+\mathrm{1}} =−\sqrt{\mathrm{2}}{i}\frac{{u}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{u}^{\mathrm{4}{n}} \\ $$$$=−\sqrt{\mathrm{2}}{i}\frac{{u}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{4}}\right)}{u}^{\mathrm{4}{n}} =−\sqrt{\mathrm{2}}{iu}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{{n}} }{u}^{\mathrm{4}{n}} \\ $$$$=−\sqrt{\mathrm{2}}{iu}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{5}}{\mathrm{4}};\:{u}^{\mathrm{4}{n}} \right)+{C}=−\sqrt{\mathrm{2}}{i}\sqrt{{x}}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{5}}{\mathrm{4}};\:{x}^{\mathrm{2}{n}} \right) \\ $$
Commented by aleks041103 last updated on 25/Dec/21
$${Yep}!\:{Thanks}\:{for}\:{finishing}\:{the}\:{problem} \\ $$