maclaurin series $\ln \left(\frac{1+x^2}{1-x^2}\right)$ I know that the Maclaurin seriess is: $\sum_{n=1}^\infty c_n(x)^n$ and $c_n = \frac{f^{n}(0)}{n!}$
$\ln \left(\frac{1+x^2}{1-x^2}\right)$
$f^0 = \ln (1+0) - \ln (1-0) = 0$
$f^{(1)} = \frac{4x}{(1+x^2)(1-x^2)} = 0$
$f^{(2)} = \frac{12x^4 + 4}{(1+x^2)^2(1-x^2)^2} = 4$
$\frac{0x^1}{1} + \frac{0x^2}{2!} + \frac{4x^2}{3!} ...$
I must have done a mistake, because I can't find the answer with that and $f^{(3)}$ seems insane.

Since the function is defined for $|x|<1$, you can write it as
$$
f(x)=\ln(1+x^2)-\ln(1-x^2)
$$
You surely know the Maclaurin series
$$
\ln(1+t)=\sum_{n\ge1}\frac{(-1)^{n+1}t^n}{n}
\qquad
\ln(1-t)=-\sum_{n\ge1}\frac{t^n}{n}
$$
and so
$$
\ln(1+x^2)=\sum_{n\ge1}\frac{(-1)^{n+1}x^{2n}}{n}
$$
and
$$
\ln(1-x^2)=-\sum_{n\ge1}\frac{x^{2n}}{n}
$$
Hence
$$
f(x)=
\sum_{n\ge1}\frac{(-1)^{n+1}x^{2n}}{n}+
\sum_{n\ge1}\frac{x^{2n}}{n}
$$
What are the terms that “survive”?

Only those with odd $n$, so you get $$\sum_{\substack{n\ge1\\n\text{ odd}}}\frac{2x^{2n}}{n}=\sum_{n\ge1}\frac{2x^{4n-2}}{2n-1}=\sum_{n\ge0}\frac{2x^{4n+2}}{2n+1}$$

