The partial fraction decomposition of $\frac{30}{(x-1)(x+1)}$ can be written in the form of $\frac{f(x)}{x-1}+\frac{g(x)}{x+1}$, where \[ \begin{array}{l} f(x)= \\ g(x)= \end{array} \]

Step 1 ：We have that \[\frac{30}{(x-1)(x+1)} = \frac{f(x)}{x-1} + \frac{g(x)}{x+1}.\]

Step 2 ：Multiplying both sides by \((x-1)(x+1)\), we get \[30 = f(x)(x+1) + g(x)(x-1).\]

Step 3 ：Setting \(x = 1\), we get \(2f(1) = 30\), so \(f(1) = 15\).

Step 4 ：Setting \(x = -1\), we get \(-2g(-1) = 30\), so \(g(-1) = -15\).

Step 5 ：Hence, \(f(x) = 15\) and \(g(x) = -15\).

Step 6 ：Finally, the partial fraction decomposition of \(\frac{30}{(x-1)(x+1)}\) is \[\frac{15}{x-1} - \frac{15}{x+1}.\]