Precalculus, Chapter 7, 7.4, Section 7.4, Problem 21

You need to decompose the fraction into irreducible fractions, such that:
(3)/(x^3+x-2) = (3)/(x^3+x-1-1) = 3/((x^3-1) + (x-1))
3/((x^3-1) + (x-1)) = 3/((x-1)(x^2+x+1) + (x-1))
3/((x-1)(x^2+ x + 2)) = A/(x-1) + (Bx+C)/(x^2+ x + 2)
You need to bring the fractions to a common denominator:
3 = Ax^2 + Ax + 2A + Bx^2 - Bx + Cx - C
You need to group the terms having the same power of x:
3 = x^2(A+B) + x(A - B + C) + 2A - C
Comparing both sides yields:
A+B =0 => A = -B
A - B + C = 0 => 2A + C = 0
2A - C = 3
Adding the relations yields:
4A =3 => A = 3/4 => B = -3/4 => C = -6/4
Hence, the partial fraction decomposition is 3/((x-1)(x^2+ x + 2)) = 3/(4x-4) + (-3x-6)/(4x^2+ 4x + 8).