Find series expansion of $f(x)=x^2\tan^{-1}(x^3)$ I was requested  to find the power expansion of $f(x)=x^2\tan^{-1}(x^3)$ and, since I'm new to power series, was wondering if my procedure is correct. I explain it by parts.
$i)$ Consider the power expansion of $\frac{d}{dx} \tan^{-1}(x^3)$:
$$\frac{d}{dx}\tan^{-1}(x^3) = \frac{3x^2}{1+x^6} = 3x^2 \sum_{n=0}^\infty (-x^6)^n = \sum_{n=0}^\infty (-1)^n \space 3x^{6n+2}$$
$ii)$ Clearly,
$$f(x) = x^2\tan^{-1}(x^3) = x^2 \int \frac{d}{dx} \tan^{-1}(x^3) dx = x^2 \int \sum_{n=0}^\infty (-1)^n \space 3x^{6n+2}dx$$
due to step $i$.
$iii)$
$$f(x)=x^2\int\sum_{n=0}^\infty (-1)^n \space 3x^{6n+2}dx = x^2 \sum_{n=0}^\infty (-1)^n \space \frac{3x^{6n+3}}{6n+3} = \sum_{n=0}^\infty (-1)^n \space \frac{3x^{6n+5}}{6n+3} = \sum_{n=0}^\infty (-1)^n \space \frac{x^{6n+5}}{2n+1}$$
$$=\frac{x^{11}}{9}  - \frac{x^{17}}{15}  + \frac{x^{23}}{21}  - \frac{x^{29}}{27}  +...   $$
Note. The series convergence radius will be $|x^6| < 1 = (-1, 1)$
Is this solution correct? I tried checking with online series expansion calculators, but they seem incapable of handling the particular problem.

$$\tan^{-1} z= \sum_{k=0}^{\infty} (-1)^k \frac{z^{2k+1}}{2k+1}$$
So $$x^2\tan^{-1} x^3=\sum_{k=0}^{\infty} (-1)^k \frac{x^{6k+5}}{2k+1}.$$
