Question

Partial fraction decomposition?
1/[x(x^2+1)]
I got as far as 1= Ax^2 + A + Bx, and when I try to write the new system, I get
A = 0
A = 1
B = 0
Which I know doesn't make sense, because A can't be both, right? What am I doing wrong?

Answer 1

if there is a quadratic in denominator, we take
cx+d/(ax^2+bx+c)

so in your case , the partial fraction will be
\(\large \dfrac{1}{x(x^2+1)}=\dfrac{A}{x}+\dfrac{Bx+c}{x^2+1}\)

now try :)

Answer 2

@MGR
whenever you come online.

Answer 3

one thing that always got me was when we had just an x^2 in a denominator, which is technically a quadratic, but it is treated as a linear factor - prolly because it is not prime

if there is a prime quadratic, we use a linear on top.

Answer 4

one neat thing to try is to factor it into its complex parts and just use constants for tops :)

Answer 5

even for x^2 we use same thing, Bx+C, but we write it different way
\(\large \dfrac{1}{x^2(x+1)}= \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2}\)
this same thing we write as
\(\large \dfrac{A}{x+1} +\dfrac{B}{x}+\dfrac{C}{x^2}\)

Answer 6

hmm, that does make sense ... i should prolly get some better textbooks :) thnx

Answer 7

Thanks, hartnn! That was a lot better xD

Answer 8

welcome ^_^