Example 4 The odd function $f(x)$ is monotonically decreasing on the interval $(-\infty, 0)$, $f(2)=0$, solve the inequality $(x-1) f(x+1)>0$.
Analyzing $f(x)$ is a function without a given analytical expression, and the properties involved are not singular, so it is necessary to use a graph to aid in thinking. For this purpose, first, based on the monotonicity of $f(x)$ and the condition $f(-2)=-f(2)=0$, make a schematic diagram of the function in the left half-plane. Then, according to the symmetry, make the graph in the right half-plane. With this figure $1-1$, we can easily see that when $x-1>0$, $x+1>2$, at this time $f(x+1)<0$, the original inequality has no solution. When $x-1<0$, due to the function being discontinuous at $x=0$, it should be divided into two cases: $x+1<0$ and $0<x+1<2$. Noting that when $0<x+1<2$, the original inequality has no solution, thus the original inequality is equivalent to $\left\{\begin{array}{l}x+1<0, \\ f(x+1)<0 .\end{array}\right.$ Then, based on the monotonicity of the function, solving it yields the solution set of the original inequality as $\{x \mid-3<x<-1\}$.