8. Define the length of intervals $(c, d),[c, d],(c, d],[c, d)$ to be $d-c$, where $d>c$. Given real numbers $a > b$, the length sum of intervals formed by $x$ that satisfy $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$ is $\qquad$ .
8. 2 Detailed Explanation: From $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$, we get $\frac{x-b+x-a-[(x-a)(x-b)]}{(x-a)(x-b)} \geqslant 0$. Simplifying, we obtain $-\frac{x^{2}-(a+b+2) x+a b+a+b}{(x-a)(x-b)} \geqslant 0$. Let $g(x)=x^{2}-(a+b+2) x+a b+a+b$, then $g(a)=b-a<0$, and $g(b)=a-b<0$. Since $a>b$, we have $b<x_{1}<a<x_{2}$.

The solution set of the inequality $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$ is $\left(b, x_{1}\right] \cup\left(a, x_{2}\right]$, and the sum of the lengths of the intervals is $\left|x_{2}-a\right|+\mid x_{1}-b \mid=\left(x_{2}+x_{1}\right)-(a+b)=2$.