To express #x/(x^3−x^2−2x+2)# in partial fractions, we should first factorize #(x^3−x^2−2x+2)#.

#(x^3−x^2−2x+2)# = #x^2(x-1)-2(x-1)=(x^2-2)(x-1)#

Hence, #x/((x^2-2)(x-1))hArr((Ax+B)/(x^2-2))+C/(x-1)#

The latter can be expanded as #((Ax+B)(x-1)+C(x^2-2))/((x^2-2)(x-1))# or

#(Ax^2+Bx-Ax-B+Cx^2-2C)/((x^2-2)(x-1))# or

#((A+C)x^2+(B-A)x-B-2C)/((x^2-2)(x-1))# and as it is equal to #x/(x^3−x^2−2x+2)#, we have

#A+C=0#, #B-A=1# and #-B-2C=0#

Eliminating C from first and third equation, by multiplying first by #2# and adding it to third, we get #2A-B=0#.

Adding this to second, we get #A=1#. This gives us #B=0# and then #C=-1#.

Hence #x/(x^3−x^2−2x+2)hArrx/(x^2-2)-1/(x-1)#