подробное решение
∫ du/(u^2-a^2)=(1/(2a))ln|(u-a)/(u+a)|+C
3x^2-9x+6=3*(x^2-3x+2)=3*((x-(3/2))^2-1/4)
∫ dx/(3x^2–9x+6)=(1/3) ∫ dx/ [b]([/b](x-(3/2))^2-1/4 [b])[/b]
u=x-3/2
du=dx
a^2=1/4
=(1/3)*(1/2*(1/2))ln|(x-(3/2)-(1/2)/(x-(3/2)+(1/2))|+C=
=(1/3)ln|(x-2)/(x-1)|+C