<s>[INST] Can you tell which are correct terms of the sum of solution of the integral $\int (x^2+1)^n dx $? According to WolframAlpha,
$$\int (x^2+1)^n dx = x \cdot _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)$$ 
and 
$$_2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{n=0}^{\infty} \frac{1}{3}(-n)\frac{(-x^2)^n}{n!}.$$
Can you tell which are correct terms of the sum of solution of the integral  $\int (x^2+1)^n dx $?
these
$$(n=0): x, (n=1): 1/3x(x^2+3), (n=2): 1/3 x(x^4+x^2), \dots$$
or (by using $_2F_1(a,b;c;z)= \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$(1)), these which I got $$(n=0): 0, (n=1): 1/3x(x^2), (n=2): 1/3 x(-x^4+x^2), \dots$$
If mine are incorrect, can you compute some terms of the sum (1)?
 [/INST] You are getting the wrong answer. The answer should be
$$ x\,{\mbox{$_2$F$_1$}\left(\frac{1}{2},-n;\,\frac{3}{2};\,-{x}^{2}\right)}. $$
Now, you won't have problems. For $n=0,1,2,3$, we get the corresponding answers 
$$ x,\,
\frac{x}{3} \left( {x}^{2}+3 \right), 
\, \frac{x}{15} \left( 3\,{x}^{4}+10\,{x}^{2}+15 \right),\, \, 
\frac{x}{35} \left( 5\,{x}^{6}+21\,{x}^{4}+35\,{x}^{2}+35 \right). $$
Added: You should use different index for the sum as
$$  _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{k=0}^{\infty} \frac{(1/2)_k(-n)_k}{(3/2)_k}\frac{(-x^2)^k}{k!}.  $$
Now, you need this identity
$$ (a)_m =  \frac{\Gamma(m+a)}{\Gamma(a)}= \frac{(m+a-1)!}{(a-1)!}.  $$
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