Calculate the sum of a function series with indefinite integral

Question

We have the series $$ f(x) = \sum\limits_{n=0}^{\infty}\frac{n}{n+2}(x^2-4)^{2n} $$ and it's asked to find the result of the series. I've tried this approach $$ \frac{n}{n+2} = \left(\frac{n+2}{n+2}-\frac{2}{n+2}\right) = 1 - \frac{2}{n+2} $$$$ y = (x^2-4)^{2}\Rightarrow \sum\limits_{n=0}^{\infty}\frac{n}{n+2}(x^2-4)^{2n} = \sum\limits_{n=0}^{\infty}\left(1 - \frac{2}{n+2}\right)y^{n} $$ then $$ \sum\limits_{n=0}^{\infty}y^{n} - \frac{2}{n+2}y^{n} = \sum\limits_{n=0}^{\infty}y^{n} - \frac{2}{y^2}\sum\limits_{n=0}^{\infty}\frac{1}{n+2}y^{n+2} $$$$ \frac{1}{1-y} - \frac{2}{y^2}\sum\limits_{n=0}^{\infty}\int y^{n+1}dy $$ which leads to $$ \frac{1}{1-y} - \frac{2}{y}\sum\limits_{n=0}^{\infty}\int y^{n}dy = \frac{1}{1-y} - \frac{2}{y}\int\left(\sum\limits_{n=0}^{\infty} y^{n}\right)dy $$ and finally to $$ \frac{1}{1-y} - \frac{2}{y}\int\frac{dy}{1-y} = \frac{1}{1-y}+\frac{2\ln|1-y|}{y} $$$$ f(x) = \frac{1}{1-(x^2-4)^{2}}+\frac{2\ln(1-(x^2-4)^{2})}{(x^2-4)^{2}} $$ where we consider values $|(x^2-4)| < 1$ and $|(x^2-4)| \neq 0$. Is this process right? I'm not sure about taking the $y$ out of the sum, and splitting the sum in two.

Answer 1

$$ \begin{align} -\sum_{n=0}^{\infty}\frac{y^n}{n+2} &= -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\frac{y^{n+2}}{n+2} = -\frac{1}{y^2}\,\sum_{n=0}^{\infty}\,\int_{\color{red}{0}}^{\color{red}{y}}t^{n+1}\,dt = -\frac{1}{y^2}\,\int_{0}^{y}\,\sum_{n=0}^{\infty}\,t^{n+1}\,dt \\[2mm] &= -\frac{1}{y^2}\,\int_{0}^{y}\frac{t}{1-t}\,dt = -\frac{1}{y^2}\,\int_{0}^{y}\frac{t\color{red}{-1+1}}{1-t}\,dt = \frac{1}{y^2}\,\int_{0}^{y}\left(1-\frac{1}{1-t}\right)dt \\[2mm] &= \frac{1}{y^2}\left[\color{white}{\frac{}{}}t+\log(1-t)\color{white}{\frac{}{}}\right]_{0}^{y}\, = \color{red}{\frac{y+\log(1-y)}{y^2}} \\[6mm] \implies f(x) &= \frac{1}{1-(x^2-4)^2}+2\,\frac{(x^2-4)^2+\log\left(1-(x^2-4)^2\right)}{(x^2-4)^4} \end{align} $$