<s>[INST] Find the series of $f(x) = x^3\arctan(x^3)$ around $x=0$ Am I correct?
$$x^3\arctan(x^3)$$ 
$$\frac{d}{dx}\arctan(x^3) = \frac{3x^2}{x^6+1} = \sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$x^3\sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$f(x) = x^3\arctan(x^3) = 3x^5\sum_{n=0}^{\infty}(x^6)^n$$
 [/INST] For $X $ such that $X^2 <1$,
$$\frac {1}{1+X^2}=\sum_{n=0}^{+\infty}(-1)^nX^{2n} $$
and by integration
$$\arctan (X)=\arctan (0)+\sum_{n=0}^{+\infty}\frac {(-1)^nX^{2n+1}}{2n+1} $$
replace $X $ by $x^3$ and you can finish and get

$$f (x)=\sum_{n=0}^{+\infty}\frac {(-1)^nx^{6n+6}}{2n+1} $$

for $x $ such that $|x|<1$.
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