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

$$\int \frac{dx}{\sqrt{x^3-4x}}$$

Answered by aleks041103 last updated on 25/Dec/21

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

Commented by aleks041103 last updated on 25/Dec/21

$$thisintegralcannotbesolvedusingelementaryfunctions$$

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

$$=-\sqrt{2}i\int \frac{du}{\sqrt{1-u^4}}=-\sqrt{2}i\underset{n=0}{\sum ^{\mathrm{\infty }}}\int \frac{{\left(\frac{1}{2}\right)}_n}{n!}{u}^{4n}du$$$$=-\sqrt{2}i\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(4n+1)}{u}^{4n+1}=-\sqrt{2}i\frac{u}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(n+\frac{1}{4})}{u}^{4n}$$$$=-\sqrt{2}i\frac{u}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n\mathrm{\Gamma }(n+\frac{1}{4})}{n!\mathrm{\Gamma }(n+\frac{5}{4})}{u}^{4n}=-\sqrt{2}iu\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n}{n!{\left(\frac{5}{4}\right)}_n}{u}^{4n}$$$$=-\sqrt{2}iu\underset{2}{}{F}_1(\frac{1}{4},\frac{1}{2};\frac{5}{4};u^{4n})+C=-\sqrt{2}i\sqrt{x}\underset{2}{}{F}_1(\frac{1}{4},\frac{1}{2};\frac{5}{4};x^{2n})$$

Commented by aleks041103 last updated on 25/Dec/21

$$Yep!Thanksforfinishingtheproblem$$

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