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Question Number 161999 by mahdipoor last updated on 25/Dec/21

∫(dx/( (√(x^3 −4x))))

$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}} \\ $$

Answered by aleks041103 last updated on 25/Dec/21

∫(dx/( (√(x^3 −4x))))=∫(dx/( (√x))) (1/( (√(x^2 −4))))= =2∫((d((√x)))/( (√(((√x))^4 −4))))=2∫((d((√x)))/( 2(√((((√x)/( (√2))))^4 −1))))= =(√2)∫(du/( (√(u^4 −1))))

$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}}=\int\frac{{dx}}{\:\sqrt{{x}}}\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}= \\ $$$$=\mathrm{2}\int\frac{{d}\left(\sqrt{{x}}\right)}{\:\sqrt{\left(\sqrt{{x}}\right)^{\mathrm{4}} −\mathrm{4}}}=\mathrm{2}\int\frac{{d}\left(\sqrt{{x}}\right)}{\:\mathrm{2}\sqrt{\left(\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{4}} −\mathrm{1}}}= \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{du}}{\:\sqrt{{u}^{\mathrm{4}} −\mathrm{1}}} \\ $$

Commented by aleks041103 last updated on 25/Dec/21

this integral cannot be solved using elementary functions

$${this}\:{integral}\:{cannot}\:{be}\:{solved}\:{using}\:{elementary}\:{functions} \\ $$

Commented by Ar Brandon last updated on 25/Dec/21

=−(√2)i∫(du/( (√(1−u^4 ))))=−(√2)iΣ_(n=0) ^∞ ∫((((1/2))_n )/(n!))u^(4n) du =−(√2)iΣ_(n=0) ^∞ ((((1/2))_n )/(n!(4n+1)))u^(4n+1) =−(√2)i(u/4)Σ_(n=0) ^∞ ((((1/2))_n )/(n!(n+(1/4))))u^(4n) =−(√2)i(u/4)Σ_(n=0) ^∞ ((((1/2))_n Γ(n+(1/4)))/(n!Γ(n+(5/4))))u^(4n) =−(√2)iuΣ_(n=0) ^∞ ((((1/2))_n ((1/4))_n )/(n!((5/4))_n ))u^(4n) =−(√2)iu _2 F_1 ((1/4), (1/2); (5/4); u^(4n) )+C=−(√2)i(√x) _2 F_1 ((1/4), (1/2); (5/4); x^(2n) )

$$=−\sqrt{\mathrm{2}}{i}\int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }}=−\sqrt{\mathrm{2}}{i}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{u}^{\mathrm{4}{n}} {du} \\ $$$$=−\sqrt{\mathrm{2}}{i}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{4}{n}+\mathrm{1}\right)}{u}^{\mathrm{4}{n}+\mathrm{1}} =−\sqrt{\mathrm{2}}{i}\frac{{u}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{u}^{\mathrm{4}{n}} \\ $$$$=−\sqrt{\mathrm{2}}{i}\frac{{u}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{4}}\right)}{u}^{\mathrm{4}{n}} =−\sqrt{\mathrm{2}}{iu}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{{n}} }{u}^{\mathrm{4}{n}} \\ $$$$=−\sqrt{\mathrm{2}}{iu}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{5}}{\mathrm{4}};\:{u}^{\mathrm{4}{n}} \right)+{C}=−\sqrt{\mathrm{2}}{i}\sqrt{{x}}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{5}}{\mathrm{4}};\:{x}^{\mathrm{2}{n}} \right) \\ $$

Commented by aleks041103 last updated on 25/Dec/21

Yep! Thanks for finishing the problem

$${Yep}!\:{Thanks}\:{for}\:{finishing}\:{the}\:{problem} \\ $$

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