Find-the-integral-of-dt-1-t-10-

Question Number 98929 by Dwaipayan Shikari last updated on 17/Jun/20

F i n d [] t h e [] i n t e g r a l [] o f []

\int \frac{d t}{\sqrt{(1 + t^{10})}}

Commented by maths mind last updated on 17/Jun/20

\frac{1}{\sqrt{1 + x}} = 1 + \sum_{n ⩾ 1} \frac{\underset{k = 1}{\prod^{n}} (\frac{1}{2} - k)}{n!} x^{n}

= 1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n)!}{4^{n} . {(n!)}^{2}} x^{n}

\frac{1}{\sqrt{1 + t^{10}}} = 1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} . 2 n!}{4^{n} . n! . n!} . x^{10 n}

\int \frac{d t}{\sqrt{1 + t^{10}}} = x (1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n)!}{4^{n} . {(n!)}^{2}} . \frac{x^{10 n}}{10 n + 1}) + c

2 n! = 4^{n} . n! . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2})

= x (1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} . 4^{n} . n! . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2})}{4^{n} . n! . 10 (n + \frac{1}{10})} . \frac{x^{10 n}}{n!}) + c

= x (1 + \sum_{n ⩾ 1} \frac{\underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2}) . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{10})}{\underset{k = 0}{\prod^{n - 1}} (k + \frac{11}{10})} . \frac{{(- x^{10})}^{n}}{n!}) + c

= x_{2} F_{1} (\frac{1}{2}, \frac{1}{10}; \frac{11}{10}; - x^{10}) + c

Commented by M±th+et+s last updated on 17/Jun/20

w e l l d o n e . i c h e c k e d i t {i t}^{'} s n i c e a n d c o r r e c t

Commented by maths mind last updated on 18/Jun/20

t h a n k y o u S i r

i c a n t f i n d y o u r p o s t \sum_{n} \sum_{m} \frac{Γ (n + \frac{1}{2}) . Γ (m + \frac{1}{2})}{Γ (n) Γ (m)} \dots \dots

c a n y o u s e n d t h e I d o f q u a t i o n i g o t a n i d e a t h a n k y o u

Commented by M±th+et+s last updated on 18/Jun/20

g o t o q 98434

Commented by M±th+et+s last updated on 18/Jun/20

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

Commented by maths mind last updated on 20/Jun/20

w i t h e P l e a s u r t h a n k y o u i w i l l p o s t s o l u t i o n

n o t y e t f i n i s h t h i s o n e