<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] Way faster:
$$ \frac{t^2}{1-t^4}=\sum_{n\geq 0} t^{4n+2} $$
for any $t\in(-1,1)$, so by termwise integration
$$ \int_{0}^{x}\frac{t^2}{1-t^4}\,dt = \sum_{n\geq 0}\frac{x^{4n+3}}{4n+3} $$
for any $x\in(-1,1)$.
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