problem with power series $\sum \frac{x^n}{n+3}$ I need to evaluate the following power series, and I don’t really know how to do it, this is the series $$\sum \frac{x^n}{n+3}$$

This is how I tackled this problem, but the final solution looks very dumb.
So we know that $$\frac{1}{1-x}=\sum  x^n$$
But we also know that we can get that $n+3$if we integrate $x^n+2$
So what we want is this series $$\sum x^{n+2}$$
Now, this series is equal to the original times $x^2$, so we can do this$$\sum x^{n+2} = x^2 \frac{1}{1-x}$$ Now we take the integral of both sides and we get something like this $$\sum \frac{x^{n+3}}{n+3} = (-\frac{x^2}{2}-1-\ln|x-1|)$$ But know we see that the left-hand side is $$x^3 \sum \frac{x^n}{n+3}$$ so we can multiply both sides by $\frac{1}{x^3}$ and we get the following $$\sum \frac{x^n}{n+3}=-\frac{1}{2x}-\frac{1}{x^3}-\frac{\ln|x-1|}{x^3}$$Can someone tell me if this answer is acceptable or if I have completely messed up?
 A: Note that
$$-\int\frac{x^2}{x-1}dx=-\int\frac{x^2-1+1}{x-1}dx=-\int\frac{(x+1)(x-1)+1}{x-1}dx$$
$$=-\int(x+1)dx-\int\frac{1}{x-1}dx=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|+C$$
So
$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{n+3}=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|$$
and we have
$$\sum_{n=0}^{\infty}\frac{x^n}{n+3}=-\frac{1}{2x}-\frac{1}{x^2}-\frac{1}{x^3}\ln|x-1|$$
and as mentioned by Gary in the comments, note that $\ln|x-1|$ can be replaced by $\ln(1-x)$ since the series converges absolutely for $|x|<1$
A: Denote
\begin{equation}
f(x)=\sum_{n=1}^\infty\frac{x^n}{n+k}, \quad k\in\{0,1,2,\dotsc\}.
\end{equation}
Then
\begin{equation}
x^kf(x)=\sum_{n=1}^\infty\frac{x^{n+k}}{n+k}
\end{equation}
and
\begin{equation}
\bigl[x^kf(x)\bigr]'=\sum_{n=1}^{\infty}x^{n+k-1}=x^k\sum_{n=0}^{\infty}x^n=\frac{x^k}{1-x}, \quad -1\le x<1.
\end{equation}
Integrating with respect to $x$ over the interval between $0$ and $t\in[-1,1)$ on both sides of the above euqation gives
\begin{equation}
t^kf(t)=\int_0^t\frac{x^k}{1-x}\textrm{d}\,x, \quad t\in[-1,1).
\end{equation}
This means that
\begin{equation}
f(t)=\begin{cases}\displaystyle
\frac{1}{t^k}\int_0^t\frac{x^k}{1-x}\textrm{d}\,x, & t\in[-1,1)\setminus\{0\};\\
0, & t=0.
\end{cases}
\end{equation}
Taking $k=3$ in the above equation, directly computing the definite integral, and simplifying yiled
\begin{equation}
f(t)=\begin{cases}
-\dfrac{1}{3}-\dfrac1{2t}-\dfrac1{t^2}-\dfrac{\ln(1-t)}{t^3}, & t\in[-1,1)\setminus\{0\};\\
0, & t=0.
\end{cases}
\end{equation}
