Ive been trying to wrap my head around binomial series to solve this problem, but Im kind of lost as to how to find the Maclaurin representation in sigma form. The rest of the problem I know how to solve:
$$\int\sqrt{1+x^3}dx$$
I know that a binomial series where $|x|$ < 1 is defined by: $$(1+x)^k = \sum_{n=0}^\infty \binom{k}{n}x^n$$
So, $$(1+x^3)^{\frac{1}{2}} = \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}x^{3n}$$
Edit: $$(1+x^3)^\frac{1}{2} = 1 + \frac{x^3}{2^11!} - \frac{1x^6}{2^22!} + \frac{1*3x^9}{2^33!} - \frac{1*3*5x^12}{2^44!} +...$$
$$= 1 + \frac{x^3}{2} + \sum_{n=2}^\infty \frac{(-1)^{n+1}[1*3*5...(2n-3)]x^{3n}}{2^nn!}$$
I think I found the Maclaurin Series, but my new question is how would one go about integrating such a thing?
I know that I can:
$$\int\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \int a_n$$
Then I can integrate $a_n$ and find radius of convergence through root or ratio test.
Maybe I could:
$$Let: a_n = \sum_{n=2}^\infty \frac{(-1)^{n+1}[1*3*5...(2n-3)]x^{3n}}{2^nn!}$$
$$\int 1 + \frac{x^3}{2} + a_n = x + \frac{x^4}{8} + \lim_{n -> \infty} \biggl|\int \frac{a_{n+1}}{a_n}dx \biggl|$$
So sorry if this is stupid and is not the way to go about this. Im ultimately unsure how to deal with the arithmetic sequence $[1*3*5...(2n-3)]$
Edit 2:
I figured out that:
$$\frac{1*3*5...(2n-3)}{n!2^n} * \frac{2*4*6...2n}{2*4*6...2n} = \frac{(2n-3)!}{[2*4*6...2n]n!2^n} = \frac{(2n-3)!}{n!2^n(n-2)!2^{n-2}} = \frac{(2n-3)!}{n!(n-2)!2^{2n-2}}$$
So then:
$$ a_n = 1 + \frac{x^3}{2} + \sum_{n=2}^\infty (-1)^{n+1}x^{3n}\frac{(2n-3)!}{n!(n-2)!2^{2n-2}} $$
So now, should I do:
$$x + \frac{x^4}{8} + \lim_{n -> \infty} \biggl|\int \frac{a_{n+1}}{a_n}dx \biggl|$$
?
Any help is appreciated, thanks!