Precalculus, Chapter 7, 7.4, Section 7.4, Problem 40

(x^2-4x+7)/[(x+1)(x^2-2x+3)]=A/(x+1)+(Bx+C)/(x^2-2x+3)

Multiply through by the LCD (x+1)(x^2-2x+3)
x^2-4x+7=A(x^2-2x+3)+(Bx+C)(x+1)
x^2-4x+7=Ax^2-2Ax+3A+Bx^2+Cx+Bx+C
x^2-4x+7=(A+B)x^2+(-2A+C+B)x+(3A+C)

Equate the coefficients of like terms. Then solve for A, B, and C.
1=A+B
-4=-2A+B+C
7=3A+C

-4=-2A+B+C
-4=-2A+(1-A)+(7-3A)
-4=-6A+8
-12=-6A
2=A

1=A+B
1=2+B
-1=B

7=3A+C
7=3(2)+C
1=C

A=2, B=-1, C=1

(x^2-4x+7)/[(x+1)(x^2-2x+3)]=2/(x+1)-(x-1)/(x^2-2x+3)