$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the 
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero. 
$$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
I keep integrating the result until I get
$$\frac{x^{n+3}}{(n+3) (n+1) (n+2)}=\frac{3 x^2}{4}-\frac{1}{2} x^2 \log (1-x)-\frac{x}{2}+x \log (1-x)-\frac{1}{2} \log (1-x)$$
I thought about multiplying by $$\frac{(n+1) (n+2)}{x^3}$$
However, the answer should be this: $$\sum _{n=0}^{\infty } \frac{x^n}{n+3}=-\frac{\log (1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2 x}$$
On the interval [-1,1), and $\frac{1}{3}$ if x=0.
Could someone please show me what I did wrong and how the steps should be to get the correct answer?

I think this is a good way of handling these series:
Let $s(x)=\sum_{n=0}^{+\infty} \frac{x^n}{n+3}$ (defined for $|x|<1$). Then
$$
x^3s(x)=\sum_{n=0}^{+\infty}\frac{x^{n+3}}{n+3},
$$
and hence
$$
(x^3s(x))'=\sum_{n=0}^{+\infty}x^{n+2}=x^2\sum_{n=0}^{+\infty}x^n=\frac{x^2}{1-x}.
$$
Integrating from $0$ to $t$, we get
$$
t^3s(t)-0^3s(0)=\int_0^t\frac{x^2}{1-x}\,dx = -t-\frac{t^2}{2}-\log(1-t),
$$
so (switching back to $x$ as variable)
$$
s(x)=-\frac{1}{x^2}-\frac{1}{2x}-\frac{\log(1-x)}{x^3}.
$$
