x-3-x-3-1-dx-

Question Number 82283 by M±th+et£s last updated on 19/Feb/20

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

Commented by MJS last updated on 20/Feb/20

{it}^{'} s impossible for me to solve this

Commented by M±th+et£s last updated on 20/Feb/20

i c a n t s o l v e i t t o s i r

i p o s t e d i t a s k i n g f o r h e l p

b u t i t h i n k t h a t i t s s p e c i a l i n t e g r a l

Answered by mind is power last updated on 20/Feb/20

H e l l o

\sqrt{1 + r} = 1 + \frac{r}{2} + \sum_{k ⩾ 2} {(- 1)}^{k - 1} . \frac{(2 k - 3)!!}{2^{k} k!} . x^{k}

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

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

= \int x^{3} (1 + \sum_{k ⩾ 1} {(- 1)}^{k - 1} \frac{(2 k - 2)!}{2^{2 k - 1} k! . (k - 1)!} . x^{3 k}) d x

= \frac{x^{4}}{4} + \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1} (2 k - 2)! 4}{2^{2 k - 1} (k - 1)! . (3 k + 4) . k!} x^{3 k + 4}

= \frac{x^{4}}{4} (1 + \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1} (2 k - 2)!}{2^{2 k - 1} (k - 1)! (3 k + 4)} . \frac{x^{3 k}}{k!})

= \frac{x^{4}}{4} (1 + \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1} . 4 . (2 k - 3)!!}{2^{k} (k + 4)} . \frac{x^{3 k}}{k!})

= \frac{x^{4}}{4} (1 + \sum_{k ⩾ 1} \frac{(- \frac{1}{2}) \dots \dots . (- \frac{1}{2} + k - 1) \frac{4}{3} \dots . . (k + \frac{4}{3} - 1)}{(\frac{7}{3}) \dots \dots (\frac{7}{3} + k - 1)} . {(\frac{x^{3}}{2})}^{k} . \frac{1}{k!}

= \frac{x^{4}}{4}_{2} F_{1} (- \frac{1}{2}, \frac{4}{3}; \frac{7}{3}; (\frac{x^{3}}{2})) + c

Commented by MJS last updated on 20/Feb/20

I thought there was a possibility using series

but I have never learned these things \dots

Commented by M±th+et£s last updated on 20/Feb/20

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