The partial fractions form of $\frac{1}{\left(x-1\right)\left(x+1\right)\left(x-2\right)}$  is

$1=A\left(x+1\right)\left(x-2\right)+B\left(x-1\right)\left(x-2\right)+C\left(x-1\right)\left(x+1\right)$

Put $x=1$  then $A=\frac{-1}{2}$  ;

put $x=-1$  then $B=\frac{1}{6}$ ;

Put $x=2$  then $C=\frac{1}{3}$

$\therefore \frac{1}{6\left(x+1\right)}-\frac{1}{2\left(x-1\right)}+\frac{1}{3\left(x-2\right)}$