Precalculus, Chapter 7, 7.4, Section 7.4, Problem 22

(x+1)/(x^2-x-6)
Let's factorize the denominator,
x^2-x-6=x^2-3x+2x-6
=x(x-3)+2(x-3)
=(x-3)(x+2)
:.(x+1)/(x^2-x-6)=(x+1)/((x-3)(x+2))
Now let(x+1)/(x^2-x-6)=A/(x-3)+B/(x+2)
(x+1)/(x^2-x-6)=(A(x+2)+B(x-3))/((x-3)(x+2))
(x+1)/(x^2-x-6)=(Ax+2A+Bx-3B)/((x-3)(x+2))
:.(x+1)=Ax+2A+Bx-3B
x+1=(A+B)x+2A-3B
Equating the coefficients of the like terms,
A+B=1
2A-3B=1
Now let's solve the above two equations to get the values of A and B,
express B in terms of A from the first equation,
B=1-A
substitute the expression of B in the second equation,
2A-3(1-A)=1
2A-3+3A=1
5A-3=1
5A=1+3
5A=4
A=4/5
Plug the value of A in the first equation,
4/5+B=1
B=1-4/5
B=1/5
:.(x+1)/(x^2-x-6)=4/(5(x-3))+1/(5(x+2))
Now let's check the above result,
RHS=4/(5(x-3))+1/(5(x+2))
=(4(x+2)+1(x-3))/(5(x-3)(x+2))
=(4x+8+x-3)/(5(x^2+2x-3x-6))
=(5x+5)/(5(x^2-x-6))
=(5(x+1))/(5(x^2-x-6))
=(x+1)/(x^2-x-6)
= LHS
Hence it is verified.