Closed form of $\sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$ 
Let
  $$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$$
  1. Find the radius of convergence.
  2. Calculate $S(x)$.
  3. Find $S^{(n)}(x)$ without computing the derivatives of $S(x)$.

From the root test I find $R = 1/4$. It's the second point that troubles me. This is my attempt:
$$\begin{align}
S(x) &= \sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n =\\
&= \sum\limits_{n=1}^\infty 2^{2n+1}\int_{-4}^x (t + 4)^{2n-1}\mathrm dt =\\
&= \int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n+1}(t + 4)^{2n-1}\mathrm dt =\\
&= 4\int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n-1}(t + 4)^{2n-1}\mathrm dt =\\
&=\ ???
\end{align}$$
I don't know how to continue from there. I know that I should transform the inner sum into a known Taylor expansion or a geometric series, but I don't see how I could do that.
As for the last point, we have that
$$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n = \sum_{n = 0}^\infty \frac{S^{(n)}(x)}{n!}(x + 4)^n,$$
as per the Taylor series definition. However, I don't know how to reconcile the indices and the two powers $2n$ and $n$.

Let $t=4(x+4)^2$ and the series becomes
$$\sum_{k=1}^\infty\frac{t^n}n.$$
You can recognize the Taylor development of $-\ln(1-t)$, or derive the series to get
$$\sum_{k=1}^\infty t^{n-1}=\frac1{1-t}.$$
