Showing that $\sum_{n=1}^{\infty}\sum_{m=1}^{x-1} \frac{m}{n^2 x^2 - m^2}=\frac{1}{2}H_{x-1}$ Consider $(1)$, $H_n$ is the nth-harmonic number

$$\sum_{n=1}^{\infty}\left({1\over (nx)^2-1}+{2\over (nx)^2-2^2}+{3\over (nx)^2-3^2}+\cdots+{x-1\over (nx)^2-(x-1)^{2}}\right)=S\tag1$$
  $x\ge2$

How does one show that $$\color{blue}{S={H_{x-1}\over 2}}?$$
An attempt:[Edited]
Because $(1)$ is the difference of two squares form, so it becomes
$$\sum_{n=1}^{\infty}\left[\left({1\over nx-1}+{1\over nx-2}+{1\over nx-3}+\cdots+{x-1\over nx-(x-1)}\right)-\left({1\over nx+1}+{1\over nx+2}+{1\over nx+3}+\cdots+{x-1\over nx+(x-1)}\right)\right]=2S\tag2$$
Not sure how to proceed next

Well, you have made a small calculation mistake. We have, $$\frac {1}{n^2x^2-1} = \frac {1}{(nx-1)(nx+1)} = \frac {1}{2}[\frac {1}{nx-1}-\frac {1}{nx+1}] $$ $$\frac {2}{n^2x^2-4} = \frac {2}{(nx-2)(nx+2)} = \frac {1}{2}[\frac {1}{nx-2}-\frac {1}{nx+2}] $$ $$\frac {3}{n^2x^2-9} = \frac {3}{(nx-3)(nx+3)} = \frac {1}{2}[\frac {1}{nx-3}-\frac {1}{nx+3}] $$ $$\vdots $$ $$\frac {x-1}{n^2x^2-(x-1)^2} = \frac {1}{2}[\frac {1}{nx-(x-1)} - \frac {1}{nx+(x-1)}] $$
Now with the numerator as $1$ in all cases, it will easily telescope for $n=1$ to $\infty $ and for $x\geq 2$, giving us, $$S = \frac{1}{2}[\frac {1}{1} + \frac {1}{2} + \frac{1}{3} + \cdots +\frac {1}{x-1}] = \frac {H_{x-1}}{2} $$
Hope it helps. 
