Precalculus, Chapter 7, 7.4, Section 7.4, Problem 26

(x+2)/(x(x^2-9))
(x+2)/(x(x^2-9))=(x+2)/(x(x+3)(x-3))
Now let (x+2)/(x(x^2-9))=A/x+B/(x+3)+C/(x-3)
(x+2)/(x(x^2-9))=(A(x+3)(x-3)+B(x)(x-3)+C(x)(x+3))/(x(x+3)(x-3))
(x+2)/(x(x^2-9))=(A(x^2-9)+B(x^2-3x)+C(x^2+3x))/(x(x^2-9))
:.(x+2)=A(x^2-9)+B(x^2-3x)+C(x^2+3x)
x+2=Ax^2-9A+Bx^2-3Bx+Cx^2+3Cx
x+2=(A+B+C)x^2+(-3B+3C)x-9A
equating the coefficients of the like terms,
A+B+C=0
-3B+3C=1
-9A=2
Solve the above three equations to get the values of A,B and C,
A=-2/9
Bach substitute the value of A in the first equation,
-2/9+B+C=0
B+C=2/9
From the above equation ,express C in terms of B
C=2/9-B
Substitute the expression of C in the second equation,
-3B+3(2/9-B)=1
-3B+2/3-3B=1
-6B=1-2/3
-6B=1/3
B=-1/18
Now plug the value of B in the expression of C,
C=2/9-(-1/18)
C=2/9+1/18
C=(2*2+1)/18
C=5/18
:.(x+2)/(x(x^2-9))=-2/(9x)-1/(18(x+3))+5/(18(x-3))
Now let's check the result,
RHS=-2/(9x)-1/(18(x+3))+5/(18(x-3))
=(-2*2(x+3)(x-3)-x(x-3)+5x(x+3))/(18x(x+3)(x-3))
=(-4(x^2-9)-(x^2-3x)+5(x^2+3x))/(18x(x^2-9))
=(-4x^2+36-x^2+3x+5x^2+15x)/(18x(x^2-9))
=(18x+36)/(18x(x^2-9))
=(18(x+2))/(18x(x^2-9))
=(x+2)/(x(x^2-9))
=LHS
Hence it is verified.