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Question Number 81400 by mind is power last updated on 12/Feb/20

H e l l o N i c e d a y i m t h i n k i n g o f t h i s o n e a c l o s e f o r m e ?

\int \sqrt{1 + x^{p}} d x

p \in R_{+},

x \in [0, 1 [

Commented by abdomathmax last updated on 13/Feb/20

a t f o r m o f s e r i e w e h a v e

{(1 + u)}^{α} = 1 + \frac{α}{1!} u + \frac{α (α - 1)}{2!} u^{2} + \dots

= 1 + \sum_{n = 1}^{\infty} \frac{α (α - 1) \dots (α - n + 1)}{n!} u^{n} \Rightarrow

\sqrt{1 + u} = 1 + \sum_{n = 1}^{\infty} \frac{\frac{1}{2} (\frac{1}{2} - 1) \dots . (\frac{1}{2} - n + 1)}{n!} u^{n} \Rightarrow

\sqrt{1 + x^{p}} = 1 + \sum_{n = 1}^{\infty} \frac{\frac{1}{2} (\frac{1}{2} - 1) \dots (\frac{1}{2} - n + 1)}{n!} x^{p n} \Rightarrow

\int \sqrt{1 + x^{p}} d x =

1 + \sum_{n = 1}^{\infty} \frac{\frac{1}{2} (\frac{1}{2} - 1) \dots (\frac{1}{2} - n + 1)}{n!} \times \frac{1}{p n + 1} x^{p n + 1} + C

Commented by mind is power last updated on 14/Feb/20

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

Commented by mind is power last updated on 14/Feb/20

n i c e S i r

Answered by behi83417@gmail.com last updated on 12/Feb/20

x^{p} = {tg}^{2} t \Rightarrow {px}^{p - 1} dx = 2 tgt (1 + {tg}^{2} t) dt

\Rightarrow dx = \frac{2}{p} {tg}^{(\frac{2}{p} - 1)} t (1 + {tg}^{2} t) dt

\Rightarrow I = \frac{2}{p} \int {tg}^{(\frac{2}{p} - 1)} t . \frac{1}{\cos^{3} t} dt =

\overset{m = (\frac{2}{p} - 1)}{=} (m + 1) \int \frac{{tg}^{m} t}{\cos^{3} t} dt =

= (m + 1) \int \sin^{m} t . \cos^{- (m + 3)} tdt = \dots

Commented by mind is power last updated on 12/Feb/20

n i c e s i r

T r y t o o u s e H y p e r g e o m e t r i c F u n c t i o n

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