Precalculus, Chapter 7, 7.4, Section 7.4, Problem 33

Decompose the denominator:
x^3-x^2-2x+2=x^2(x-1)-2(x-1)=(x^2-2)(x-1)=(x-1)(x-sqrt(2))(x+sqrt(2)).
Therefore the fraction decomposition has the form
x/(x^3-x^2-2x+2)=A/(x-1)+B/(x-sqrt(2))+C/(x+sqrt(2)).
To find A,B and C multiply both sides by the original denominator:
x=A(x^2-2)+B(x-1)(x+sqrt(2))+C(x-1)(x-sqrt(2)), or
x=x^2(A+B+C)+x(B(sqrt(2)-1)-C(sqrt(2)+1))+(-2A-Bsqrt(2)+Csqrt(2)).
Thus A+B+C=0, B(sqrt(2)-1)-C(sqrt(2)+1)=1 and -2A-Bsqrt(2)+Csqrt(2)=0.
A=-(B+C),
B(sqrt(2)-1)-C(sqrt(2)+1)=1,
2(B+C)-Bsqrt(2)+Csqrt(2)=0, orB(2-sqrt(2))+C(2+sqrt(2))=0, orB(sqrt(2)-1)+C(sqrt(2)+1)=0.
Add and subtract these two equations and obtain
2B(sqrt(2)-1)=1, or B=1/(2(sqrt(2)-1))=(sqrt(2)+1)/2 and
2C(sqrt(2)+1)=-1, or C=-1/(2(sqrt(2)+1))=-(sqrt(2)-1)/2.
And A= -(B+C)=-1.

Now check this result:
-1/(x-1)+(sqrt(2)+1)/2 1/(x-sqrt(2)) -(sqrt(2)-1)/2 1/(x+sqrt(2))=
=-1/(x-1)+1/2 ((sqrt(2)+1)x+sqrt(2)(sqrt(2)+1)-(sqrt(2)-1)x+sqrt(2)(sqrt(2)-1))/(x^2-2)=
=-1/(x-1) +1/2 (2x+4)/(x^2-2)=-1/(x-1)+(x+2)/(x^2-2)=
=(2-x^2+x^2+x-2)/((x+2)(x^2-2))=x/(x^3-x^2-2x+2),
which is correct.