How would one go by manipulating known generating functions to get the following: $$\sum_0^\infty(n+1)(2n+1)x^n$$
I know the following, which is what I am assuming I must manipulate. I have the answer to the closed form, but I do not understand how to get there. Please, no answers as of yet, just hints/tips.
$$ \sum_0^\infty(2n+1)x^{2n}=\frac{1+x^2}{(1-x^2)^2} $$
 A: Hint
Since you have $(n+1)(2n+1)x^n$ I suppose it is natural to consider that first and second derivatives, as well as function, would appear in the result.  
Since $(n+1)(2n+1)=2n^2+3n+1=2n(n-1)+5n+1$, we can then rewrite $$\sum_0^\infty(n+1)(2n+1)x^n=2x^2 \sum_0^\infty n(n-1)x^{n-2}+ 5 x \sum_0^\infty n x^{n-1}+ \sum_0^\infty  x^{n}$$ where appear $$\sum_0^\infty  x^{n}=\frac{1}{1-x}$$ as well as its first and second derivative with respect to $x$.
I am sure that you can take from here and finish.
A: The general recipe is if $f(x) = \sum\limits_{n=0}^\infty a_n x^n$, then 
$$
(x\frac{d}{dx})^kf(x) = \sum_{n=0}^{\infty} n^ka_n x^n
\quad\text{ and }\quad
\frac{1}{x^k}\int_0^x t^{k-1} f(t) dt = \sum_{n=0}^\infty \frac{a_n}{n+k}x^n
$$
If you have a sequence whose coefficients are coefficients of something function $f(x)$ you know times some polynomial in $n$, then you can obtain the generating function by applying
corresponding polynomial in $(x\frac{d}{dx})$ to $f(x)$. e.g.
$$\begin{align}
\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \quad\implies\quad
\sum_{n=0}^{\infty} (n+1)(2n+1)x^n
= & (x\frac{d}{dx} + 1)(2x\frac{d}{dx}+1)\frac{1}
{1-x}\\
= & (x\frac{d}{dx} + 1)\frac{1+x}{(1-x)^2}\\
= & \frac{1+3x}{(1-x)^3}
\end{align}
$$
A: Try the following strategy: 
step 1: replace x^2 by y on both sides.
step 2: multiply both sides by y.
step 3: differentiate both sides with respect to y.
step 4: replace y by x again and finish.
