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Question Number 86454 by redmiiuser last updated on 28/Mar/20

p l s c h e c k t h e

q u e s t i o n b e l o w

Answered by redmiiuser last updated on 28/Mar/20

\int 1 / \sqrt{} (1 + x^{3}) . d x

{(1 + x^{3})}^{(- 1 / 2)}

= 1 + (- 1 / 2) x^{3} + (- 1 / 2) (- 3 / 2) x^{6} / 2! + (- 1 / 2) (- 3 / 2) (- 5 / 2) x^{9} / 3! + (- 1 / 2) (- 3 / 2) (- 5 / 2) (- 7 / 2) / 4! x^{12} + \dots

t h e r e f o r e

{(1 + x^{3})}^{(- 1 / 2)}

= \underset{n = 0}{\sum^{\infty}} (({(- 1)}^{n} . (2 n)! . {(x^{3})}^{n}) / ((2^{2 n} . {(n!)}^{2})))

∴ \int {(1 + x^{3})}^{(- 1 / 2)} . d x

= \int \underset{n = 0}{\sum^{\infty}} (({(- 1)}^{n} . (2 n)! . {(x^{3})}^{n}) / (2^{2 n} . {(n!)}^{2})) . d x

= \underset{n = 0}{\sum^{\infty}} (({(- 1)}^{n} . (2 n)! . {(x)}^{3 n + 1}) / ((3 n + 1) . (2^{2 n}) . {(n!)}^{2}))

Commented by redmiiuser last updated on 29/Mar/20

c a n a n y o n e c o m m e n t

w h e t h e r t h e a b o v e

p r o c e s s i s c o r r e c t o r

w r o n g .

Commented by redmiiuser last updated on 29/Mar/20

m r . T a n m a y c a n y o u

h e l p m e i n t h e

a b o v e p r o b l e m p l s .

Commented by TANMAY PANACEA. last updated on 29/Mar/20

\int \frac{d x}{\sqrt{1 + x^{3}}}

x^{3} = {t a n}^{2} α

x = {(t a n α)}^{\frac{2}{3}} \to d x = \frac{2}{3} \times {(t a n α)}^{\frac{2}{3} - 1} \times {s e c}^{2} α d α

\int \frac{2 {(t a n α)}^{\frac{- 1}{3}} \times {s e c}^{2} α \times d α}{3 \times {(1 + {t a n}^{2} α)}^{\frac{1}{2}}}

\frac{2}{3} \int \frac{s e c α d α}{{(t a n α)}^{\frac{1}{3}}}

\frac{2}{3} \int \frac{d α}{{(s i n α)}^{\frac{1}{3}} {(c o s α)}^{\frac{2}{3}}}