Suppose that, #I=int(2x-1)/{x(x-2)(x-3)}dx#.
#:. I=int{(2x)/{x(x-2)(x-3)}+(-1)/{x(x-2)(x-3)}}dx#,
#=int{2/{(x-2)(x-3)}+{(x-3)-(x-2)}/{x(x-2)(x-3)}}dx#,
#=int{2/{(x-2)(x-3)}+(x-3)/{x(x-2)(x-3)}-(x-2)/{x(x-2)(x-3)}}dx#,
#=2int1/{(x-2)(x-3)}dx+int1/{x(x-2)}dx-1/{x(x-3)}dx#,
#=2int{(x-2)-(x-3)}/{(x-2)(x-3)}dx+1/2int{x-(x-2)}/{x(x-2)}dx#
#-1/3int{x-(x-3)}/{x(x-3)}dx#,
#=2int{1/(x-3)-1/(x-2)}dx+1/2int{1/(x-2)-1/x}dx#
#-1/3int{1/(x-3)-1/x}dx#,
#=2ln|(x-3)|-2ln|(x-2)|+1/2ln|(x-2)|-1/2ln|x|-1/3ln|(x-3)|+1/3ln|x|#,
#=(2-1/3)ln|(x-3)|-(2-1/2)ln|(x-2)|-(1/2-1/3)ln|x|#.
# rArr I=5/3ln|(x-3)|-3/2ln|(x-2)|-1/6ln|x|+C, or, #
# I=ln|(x-3)^(5/3)/{(x-2)^(3/2)x^(1/6)}|+C#.