Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$ 
We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$


Starting from the second expression I get :
$$ f(x) = 3x^2 - (x - 1)^2 \ln(1 - x) - (x + 1)^2 \ln(1 + x) $$
Using power series of $\ln(1 + x)$ and $ \ln(1 - x)$ does not lead to the result wanted.
I do not know how to proceed to get the first expression.
 A: Partial fraction decomposition of the coefficient yields to
$$\frac1{n(n+1)(2n+1)}=\frac1n+\frac1{n+1}-\frac4{2n+1}$$
Therefore we may write the original series as
$$\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=\sum_{n=1}^\infty\left[\frac1n+\frac1{n+1}-\frac4{2n+1}\right]x^{2n+2}$$
The easiest way from hereon is to recognize well-known series representations. To be precise we can further conclude that
\begin{align*}
&(1)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n}=x^2\sum_{n=1}^\infty\frac{x^{2n}}{n}=x^2(-\log(1-x^2))\\
&(2)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)+x^2\\
&(3)&&-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)-x\right]
\end{align*}
Now combining these three series we can easily see that

$$\therefore~f(x)=\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)$$

Of course, as HAMIDINE SOUMARE pointed out within his answer, we have to check beforehand whether these series converge or not and therefore if we are even allowed to split up the sum like this. However, I will leave this to you.

EDIT
Important to notice is that the given logarithm series starts at $n=0$ and not at $n=1$. Thus, we have to consider the first summand extra. Moreover I realised that I made two contrary mistakes which overall did not affect the solution. To be precise my given equations $(2)$ and $(3)$ both contain an error I which I will correct now. For $(2)$ it should rather be
$$(2)~~~\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)\color{red}{-x^2}$$
Hence the series expansion of $\log(1-x^2)$ is given by
$$\log(1-x^2)=-\sum_{\color{blue}{n=1}}^\infty \frac{x^{2n}}n=-x^2-\sum_{n=2}^\infty \frac{x^{2n}}n\implies \sum_{n=2}^\infty \frac{x^{2n}}n=-\log(1-x^2)-x^2$$
From hereon I could clear my doubts concerning the last sum. The sum should correctly be given by
$$(3)~~~-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)\color{red}{-2x}\right]$$
Rather easy this can justified by exploiting the series expansion of $\log\left(\frac{1+x}{1-x}\right)$, which the same as $2\arctan(x)$, given by
$$\log\left(\frac{1+x}{1-x}\right)=2\sum_{\color{blue}{n=0}}^\infty\frac{x^{2n+1}}{2n+1}=2x+2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\implies 2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}=\log\left(\frac{1+x}{1-x}\right)-2x$$
Now note that by adding up these three sums we obtain a closed-form expression of the original sum, to  be precise
\begin{align*}
f(x)&=\underbrace{[x^2(-\log(1-x^2))]}_{(1)}+\underbrace{[-\log(1-x^2)-x^2]}_{(2)}+\underbrace{\left[-2x\left[\log\left(\frac{1+x}{1-x}\right)-2x\right]\right]}_{(3)}\\
&=[-x^2+4x^2]+[-x^2\log(1-x^2)-\log(1-x^2)]+\left[-2x\log\left(\frac{1+x}{1-x}\right)\right]\\
&=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)
\end{align*}
Which is, in fact, the desired solution this time correctly derived.
A: Hint: $\frac{1}{n(n+1)(2n+1}=\frac{a}{n}+\frac{b}{n+1}+\frac{c}{2n+1}$. Find $a$, $b$ and $c$. Rewrite your series as sum of such series, justify convergence of those series and compute to get the needed function.
Recall: $\sum_{n=1}^{\infty}{\frac{t^n}{n}}=\log(1-t), \forall t, |t|<1$
