Why we keep continuing to write terms in partical fraction for integrals?

Question

For example: $\int \frac{dx}{x^4.(x+4)}$ and we rewrite integrant as $\frac {1}{x^4(x+4)}=\frac{A}{x} + \frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+4}$

My question is why it is not enough to write just $\frac{D}{x^4}$. Why we need to other terms $\frac {1}{x^4(x+4)}=\frac{A}{x} + \frac{B}{x^2}+\frac{C}{x^3}$

And also, can we use following to rewrite integrant : $\frac {1}{x^4(x+4)}=\frac{Ax^3+Bx^2+Cx+D}{x^4} +\frac{E}{x+4}$

You cannot write $\dfrac {1}{x^4(x+4)}$ as $\dfrac{D}{x^4}$ (with a constant factor $D$). – The purpose of the partial fraction decomposition is to write a rational function as a sum of “simple” fractions which are easy to integrate. — Martin R, May 25 '23 at 12:51
The point of the question is "why can't I throw away the lower powers of the repeated factor?" not "why is just $D/x^4$ not sufficient?". The latter is obviously a non-starter, but the former may look like it has a chance even though it doesn't. — Ian, May 25 '23 at 13:11
Also see Why partial fraction decomposition of $\frac{1}{s^2(s+2)}$ is $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$? and Derivation of the general forms of partial fractions and Why do you need two fractions for partial fraction decomposition with repeated factors? and many others you will find in the "Linked" or "Related" lists of those questions. — David K, May 25 '23 at 13:16
As you will find in the answers to some of the linked questions, in fact you can write $\frac {1}{x^4(x+4)}=\frac{Ax^3+Bx^2+Cx+D}{x^4} +\frac{E}{x+4}$ and that is where the extra terms such as $\frac Ax$ come from. — David K, May 25 '23 at 13:18

score 2 · Answer 1 · answered May 25 '23 at 13:08

2

You ask:

Why we need to other terms

Well, a partial fraction decomposition is similar to, in some sense, writing an expression as a sum of fractions - For example when we write:

$\frac{6}{9}=\frac{1}{9}\:+\frac{2}{9}\:+\frac{3}{9}$

We can't drop any of the fractions on the R.H.S, otherwise, the equality between the two sides would be lost.

In your case, unless $A,B,C,E$ have zero value, we can't drop them and just use the fraction with $D$ (assuming it is not zero), otherwise there would be no equality between the R.H.S and the L.H.S

The fact that the Algebraic expression on the L.H.S equals the Algebraic expression on the R.H.S must always be maintained, whether we are performing integration, or doing any other manipulation requiring equality to be preserved.

answered May 25 '23 at 13:08

NoChance

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2

I think a better example would have more factors in the denominator; e.g. $$\frac{13}{2^2\cdot3}=\frac{13}{12}=\frac34+\frac13=\frac12+\frac1{2^2}+\frac13$$ – mr_e_man May 25 '23 at 13:15
Your example is good indeed. – NoChance May 25 '23 at 13:22
But is there actually a form of PFD for integers rather than polynomials? – mr_e_man May 25 '23 at 13:52
1

Found something at the end of this answer – mr_e_man May 25 '23 at 14:14
@mr_e_man, I can't answer your question because I don't know. Thanks for the link. Its a bit involved and I will need to study it. – NoChance May 25 '23 at 20:05

Why we keep continuing to write terms in partical fraction for integrals?

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