<s>[INST] deriving the sum of $x^n/(n+2)^2$ I am writing a research paper and I have stumbled upon an issue.    
I have to evaluate
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2}$$
Here is what I did:
$$ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$
$$\sum_{n=1}^{\infty} x^{n+1} = \frac{x^2}{1-x}$$
Integrate once with respect to $x$.
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{n+2} = \frac{3}{2} - \frac{x^2}{2} - x - \log(1-x)$$
Divide by $x$
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+2} = \frac{3}{2x} - \frac{x}{2} - 1 - \frac{\log(1-x)}{x}$$
Integrate the expression again;    
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{(n+2)^2} =  \frac{3\log(x)}{2} - x - \frac{x^2}{4} + Li_2(x)$$
All we have to do is divide by $x^2$
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{3\log(x)}{2x^2} - \frac{1}{x} - \frac{1}{4} + \frac{Li_2(x)}{x^2}  = \frac{4Li_2(x) - x^2 - 4x + 6\log(x)}{4x^2}$$
The issue is WolframAlpha returns it as:
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} =  \frac{4Li_2(x) - x^2 - 4x}{4x^2}$$
Why is the way I did, wrong? I have an extraneous, extra log term.? 
Help is appreciated.
 [/INST] Not sure what you did, but all that work seems kinda silly to arrive at your result.
$$\sum_{n=1}^\infty\frac{x^n}{(n+2)^2}=\frac{1}{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{(n+2)^2}=\frac{1}{x^2}(-x-\frac{x^2}{4}+\sum_{n=1}^\infty\frac{x^n}{n^2})=\frac{1}{x^2}(-x-\frac{x^2}{4}+\text{Li}_2(x))$$
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