dx-x-3-4x-

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

\int \frac{d x}{\sqrt{x^{3} - 4 x}}

Answered by aleks041103 last updated on 25/Dec/21

\int \frac{d x}{\sqrt{x^{3} - 4 x}} = \int \frac{d x}{\sqrt{x}} \frac{1}{\sqrt{x^{2} - 4}} =

= 2 \int \frac{d (\sqrt{x})}{\sqrt{{(\sqrt{x})}^{4} - 4}} = 2 \int \frac{d (\sqrt{x})}{2 \sqrt{{(\frac{\sqrt{x}}{\sqrt{2}})}^{4} - 1}} =

= \sqrt{2} \int \frac{d u}{\sqrt{u^{4} - 1}}

Commented by aleks041103 last updated on 25/Dec/21

t h i s i n t e g r a l c a n n o t b e s o l v e d u s i n g e l e m e n t a r y f u n c t i o n s

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

= - \sqrt{2} i \int \frac{d u}{\sqrt{1 - u^{4}}} = - \sqrt{2} i \underset{n = 0}{\sum^{\infty}} \int \frac{{(\frac{1}{2})}_{n}}{n!} u^{4 n} d u

= - \sqrt{2} i \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n! (4 n + 1)} u^{4 n + 1} = - \sqrt{2} i \frac{u}{4} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n! (n + \frac{1}{4})} u^{4 n}

= - \sqrt{2} i \frac{u}{4} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n} Γ (n + \frac{1}{4})}{n! Γ (n + \frac{5}{4})} u^{4 n} = - \sqrt{2} i u \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n} {(\frac{1}{4})}_{n}}{n! {(\frac{5}{4})}_{n}} u^{4 n}

= - \sqrt{2} i u \underset{2}{} F_{1} (\frac{1}{4}, \frac{1}{2}; \frac{5}{4}; u^{4 n}) + C = - \sqrt{2} i \sqrt{x} \underset{2}{} F_{1} (\frac{1}{4}, \frac{1}{2}; \frac{5}{4}; x^{2 n})

Commented by aleks041103 last updated on 25/Dec/21

Y e p! T h a n k s f o r f i n i s h i n g t h e p r o b l e m