<s>[INST] 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?
 [/INST] 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 $|x|<1$
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