Expert: Paul Klarreich - 10/10/2010

Question
hello sir, the question is : express the following in partial fractions
1/(x^2+a)(x^2+b). i expressed this as1=(Ax+B)(x^2+b) + (Bx+C)(x^2+a). i
thrn timsed it out to give 1=(A+B)x^3 +(B+C)x^2+(Ab+Ba)x+(Bb+Ca).
equating the coefficients i got four equations...
A+B=0
B+C=o
Ab+Ba=0
Bb+Ca=1
im not sure what to do now...any help will be greatly appreciated!

Answer
Questioner:   jay
Country:  United Kingdom
Category:  Calculus
Private:  No
 
Subject:  partial fractions
Question:  hello sir, the question is : express the following in partial fractions
1/(x^2+a)(x^2+b). i expressed this as1=(Ax+B)(x^2+b) + (Bx+C)(x^2+a). i
thrn timsed it out to give 1=(A+B)x^3 +(B+C)x^2+(Ab+Ba)x+(Bb+Ca).
equating the coefficients i got four equations...
A+B=0
B+C=o
Ab+Ba=0
Bb+Ca=1
im not sure what to do now...any help will be greatly appreciated!
....................................
I don't think that is correct. I think you want this:
 
      1           Ax + B   Cx + D
--------------  = ------- + --------
(x^2+a)(x^2+b)    (x^2+a)   (x^2+b)

Now:

1  = (Ax + B)(x^2+b) + (Cx + D)(x^2+a)

1  = Ax(x^2+b) + B(x^2+b) + Cx(x^2+a) + D(x^2+a)

1  = Ax^3+ Axb + Bx^2+Bb + Cx^3+Cxa + Dx^2 + Da

Bb + Da = 1  (const)
Ab + Ca = 0  (x)
B + D = 0 (x^2)
A + C = 0    (cube)

Those are your equations.

Start by putting  D = - B,  C = - A into the first two, then solving for A, B.  You'll get it.