Theory of Partial Fraction Decomposition

Question

The function in question that I want to decompose is $$\dfrac{8x^3 + 7}{(x+1)(2x+1)^3}$$

I had the idea to to break this down into:
$$\dfrac{A}{x+1} + \dfrac{Bx^2 +Cx + D}{(2x+1)^3} + \dfrac{Ex + F}{(2x+1)^2} + \dfrac{G}{2x+1}$$

Well this turns into a really messy system of 4 equations and 7 variables. Is there an easier way to decompose the function? Can I possibly eliminate any of these variables at the outset?

Here is a partial fraction technique which handles this problem. — Mhenni Benghorbal, Jul 06 '13 at 09:27

score 3 · Accepted Answer · answered Feb 15 '13 at 18:26

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According to Partial Fraction Decomposition rule, it will be $$\frac{A}{x+1} + \frac{B}{(2x+1)^3} + \frac{E}{(2x+1)^2} + \frac{G}{2x+1}$$

answered Feb 15 '13 at 18:26

lab bhattacharjee

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@Mark, could you have a look into the link in the answer or http://en.wikipedia.org/wiki/Partial_fraction#Illustration – lab bhattacharjee Feb 15 '13 at 18:33

score 2 · Answer 2 · answered Feb 15 '13 at 19:00

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I'll simply alert you to the existence of two separate methods for dealing with partial fractions with repeated roots that may potentially simplify the process:

1) Ostrogradsky's method

2) Hardy's Method (Page 10 on)

answered Feb 15 '13 at 19:00

sponsoredwalk

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score 1 · Answer 3 · answered Feb 15 '13 at 18:28

You only need to break it down into $\displaystyle\frac{A}{x+1}$ and $\displaystyle\frac{Bx^2 + Cx + D}{(2x+1)^3}.$ What you end up with is $\begin{align*} 8A + B &= 8\\ 12A + B +C &= 0\\ 6A + C + D &= 0\\ A + D &= 7 \end{align*}$

Solving gives $A = 1, B = 0, C = -12, D =6.$

Note that this is equivalent to lab bhattacharjee's decomposition.

Theory of Partial Fraction Decomposition

3 Answers3