Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.
I would love some help with this.

The series appears to take the form
$$ f(x) = 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n $$
which can be seen in the form:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n} \, x^n - \sum_{n=1}^{\infty} \frac{(n+1) \, x^n}{n} \\
&= 2 + \frac{d^2}{d x^2} \, \sum_{n=1}^{\infty} \frac{x^{n+2}}{n} - \frac{d}{dx} \, \sum_{n=1}^{\infty} \frac{x^{n+1}}{n} \\
&= 2 + \frac{d^2}{d x^2} \, (- x^2 \, \ln(1-x) ) - \frac{d}{d x} \, (- x \, \ln(1-x) ) \\
&= \cdots \\
&= 2 + \frac{3 x - 2 x^2}{(1-x)^2} - \ln(1-x) \\
&= \frac{2 - x}{(1-x)^2} - \ln(1-x).
\end{align}
A faster way is:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \frac{n (n+2) + 1}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \left( (n+2) + \frac{1}{n} \right) \, x^n \\
&= \sum_{n=0}^{\infty} (n+2) \, x^n + \sum_{n=1}^{\infty} \frac{x^n}{n} \\
&= \frac{2-x}{(1-x)^2} - \ln(1-x).
\end{align}
