Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+2)}x^{n-1}$ for $|x|<1$.
Let $S_2(x)=x^3S_1'(x)$. Then $S_2'(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\frac{x^2}{1+x}$.
By integration, I obtained $S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$. Then how to obtain $S(x)$? Or there is other method to do this problem?
 A: $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}~~~(1)$$
Multipluy by $x$ on both sides and integrate w.r.t. $x$
$$\int x \log(1+x) dx= \sum_{n=1}^{\infty} (-1)^{n-1}\int \frac{x^{n+1}}{n} dx= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+2}}{n(n+2)} .$$
$$\implies S(x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n(n+2)}= \frac{1}{x^3} \int x \log(1+x) ~dx.$$
Finally by carrying out integration by parts, we get the required sum
$$S(x)=\frac{1}{4x^3}[2x-x^2-2\log(1+x)+2x^2 \log(1+x)]$$
A: @ Tao X : Your method is not the simplest. But what you did is correct :
 $$S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$$
Then integrate :
$$S_1(x)=\int S_1'(x)dx=\frac{1}{2x}-\frac{\ln (x+1)}{2x^2}+\frac12\ln(x+1)+C$$
We know that $S_1(0)=0$. This allows to find $C$. Expand $\ln(x+1)$ to series around $x=0$ :
$\frac{x-\ln (x+1)}{2x^2}+\frac12\ln(x+1)=\frac{x-(x-\frac12 x^2+...)}{2x^2}+\frac12(x-\frac12 x^2+...)=\frac14+\epsilon(x)$ with $\epsilon(0)=0$.
Thus $C=-\frac14$ and $S_1(x)=\frac{1}{2x}-\frac{\ln (x+1)}{2x^2}+\frac12\ln(x+1)-\frac14$ .
$$S(x)=\frac{1}{2x^2}-\frac{\ln (x+1)}{2x^3}+\frac{1}{2x}\ln(x+1)-\frac{1}{4x}$$ .
