# Integrating via partial fractions

My question is this:

Why is it that fractions have to be split up in a very specific manner?

For example if I have $\frac{5x}{(x+1)^2}$ this fraction HAS to be split up like this:$$\frac A{x+1} + \frac B{(x+1)^2}$$

And if it were $\frac{5x}{(x^2+1)^2}$ then it is split up in this way: $$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$$

So why does it have to follow this very specific pattern? I can't find any information which is relevant on the internet. Thanks.

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When doing partial fraction decomposition, each denominator in the sum $\sum_{i=1}^{n}\frac{p_i(x)}{q_i(x)}$ must be a factor of the original denominator of your function $f(x) = \frac{p(x)}{q(x)}$ (in lowest terms), so that when we write the sum with the lowerst common denominator
$$f(x) =\frac{\sum_{i=1}^{n}p_i(x)\prod_{j\neq i}q_j(x)}{q_1(x)\ldots q_n(x)}$$ we have that the resulting denominator $q_1\ldots q_n$ is divisible by the original denominator $q$. If our new denominator did not divide the original one, then it could not possibly be equal to $f(x)$ (for $p/q$ and our sum would have different denominators in lowest terms).
It is possible to use extra fraction terms whose denominators are not factors of the original denominator $q(x)$, but they will always cancel out in the end and so it is sufficient to ignore them.
The numerator can, in fact, have any degree: $p_i(x) = a_0 + a_1x+x_2x^2+\ldots + a_nx^n$. It turns out, however, that the degree of the numerators in your decomposition is limited by the degree of the numerator in the original rational function. When we take the sum of our decomposition, the resulting rational function must have (degree of denom)-(degree of num) equal to that of the original rational function. So we end up with a polynomial of finite degree. – Alqatrkapa Jun 27 '14 at 18:53
It's a matter of necessity. It's a matter of convenience that terms like $Cx^2$ are left off. – Alqatrkapa Jun 27 '14 at 21:30