<s>[INST] Find the power series in powers of $x$ for the function Find the power series in powers of $x$ for the function
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt$$
Solution Verification Requesteed
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt \implies f'(x) = \frac{x^2}{1-x^4}$$
Now, we proceed with partial fraction decomposition. We obtain
$$f'(x) = \frac{x^2}{1-x^4} = - \frac{x^2}{x^4-1} = -x^2 \cdot \frac{1}{(x^2+1)(x+1)(-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+1} + \frac{D}{x-1}$$
We do partial fractions and obtain $A=0, B= -\frac{1}{2}, C = \frac{1}{4}, D = - \frac{1}{4}$. Now, we see that
$$-x^2 \cdot \frac{1}{x^4-1} = - \frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} - \frac{1}{4(x-1)}$$
$$f'(x) = \frac{x^2}{x^4-1} =  \frac{1}{2(x^2+1)} - \frac{1}{4(x+1)} + \frac{1}{4(x-1)}$$
$$f'(x)= \frac{1}{2} \cdot \frac{1}{1-(-x^2)} - \frac{1}{4} \cdot \frac{1}{1-(-x)} + \frac{1}{4} \cdot \frac{1}{-(1-x)}$$
$$f'(x)= \frac{1}{2} \sum_{n=0}^\infty (-1)^nx^{2n} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n x^n - \frac{1}{4} \sum_{n=0}^\infty x^n$$
We can integrate each sum term-by-term to obtain
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - \frac{1}{4} \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} $$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} - \frac{1}{4} \Big[ \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}  + \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}\Big]$$
$$f(x) =  \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + \Big[(x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) + (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4})  \Big]$$
$$f(x) = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} + 2 \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$$
I'd simplify it further into one series but that dang $(-1)^n$ in the first sum keeps things from being nice and intuitive.

UPDATE
$$f'(x) = x^2 \cdot \frac{1}{1-(x^4)} = x^2 \sum_{n=0}^\infty x^{4n} = \sum_{n=0}^\infty x^{4n+2}$$
$$\implies f(x) = \sum_{n=0}^\infty \frac{x^{4n+3}}{4n+3}$$ as Jack D'Aurizio has concluded.
 [/INST] You can also proceed by identification:
$\begin{align}(1-x^4)f'(x)
&=(1-x^4)\sum\limits_{n=1}^\infty na_nx^{n-1}\\
&=\sum\limits_{n=1}^\infty na_nx^{n-1}-\sum\limits_{n=1}^\infty na_nx^{n+3}\\
&=\sum\limits_{n=0}^\infty (n+1)a_{n+1}x^n-\sum\limits_{n=4}^\infty (n-3)a_{n-3}x^n\\
&=a_1+2a_2x+3a_3x^2+4a_4x^3+\sum\limits_{n=4}^\infty \Big((n+1)a_{n+1}-(n-3)a_{n-3}\Big)x^n\\
&=x^2
\end{align}$
Leads to the system:
$\begin{cases}
a_1=0\\
a_2=0\\
a_3=\frac 13\\
a_4=0\\
na_n=(n-4)a_{n-4}&\forall n\ge 5\end{cases}$
Also $a_0=f(0)=0$
Therefore all coefficients are zero except for $i=4n+3$ which verifies the equation $i\times a_i=1$
(i.e. $3a_3=1$ propagates to $7a_7=(7-4)a_3=3a_3=1$ and so on).

$$f(x)=\sum\limits_{n=0}^\infty \frac{x^{4n+3}}{4n+3}$$

</s>