Integration of high order fraction; explanation of method.

Question

I understand the cases where the order is less or equal to 3 (example where it is three we split numerator with A,B,C), but in this case(Example 8) I do not see why we split the numerator with A, Bx+C, Dx+E, why do we add x, and not use A B C, D, E alone. I guess what I am asking is how to find the partial fraction decomposition.

Answer 1

There are two different things going on here. First there is the issue that when a factor is repeated, we wind up needing a term for it with each degree up to the number of repetitions. Second there is the issue that when we have an irreducible quadratic factor, its numerator needs to be linear instead of constant.

Both of these basically come down to "you can't represent the numerator in the right way otherwise". Let's start by looking at a simpler example of the second problem. Consider decomposing

$$\frac{2x^2+1}{x^3+x}.$$

You factor the denominator and you get $x(x^2+1)$. Suppose you now try to write the numerator as $Ax+B(x^2+1)$. You can't do it; to get the quadratic coefficient and the constant coefficient to both match, you would need $B$ to be both $1$ and $2$, which is impossible. In fact, in terms of counting variables, this should make sense, because the numerator could be any quadratic, which is defined in terms of three coefficients, but this form only has two unknown coefficients.

If you instead try to write the numerator as $(Ax+B)x+C(x^2+1)$, then you can do it: you now have three variables and three equations, namely $A+C=2$, $B=0$, $C=1$ (going through terms of degree two, one, and zero).

The problem is very similar with repeated factors. Again let's try to decompose

$$\frac{2x^2+2x+1}{x^2(x+1)}.$$

Again if you try to write the numerator as $A(x+1)+Bx^2$, you can't do it: you'd need $A=2$ and $A=1$ at the same time. But if you write it as $A(x+1) + Bx + Cx^2$, all works out. As in the previous case, the numerator could be any quadratic (three coefficients), so you need three coefficients in order for everything to match.

A more sneaky way to see this is to consider how I made these examples in the first place: I started from $\frac{1}{x} + \frac{x}{x^2+1}$ and $\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}$ and then "undid" the partial fraction decomposition. They still wind up in the form where the degree of the numerator is less than the degree of the denominator, so there must be more involved than in the simple cases.

Answer 2

Simply, not all functions can be written in the form indicated without the $x$ terms in the numerator. For example, consider the expression $$\frac{x}{x^2 + 1}.$$ Certainly this cannot be written in the form $$\frac{A}{x^2 + 1}:$$ cross-multiplying would give $A = x$, but $A$ is a constant.

Alternatively, and practically, we can see that, even for the above example, without terms in $x$, there would be no solution to the given system in the coefficients $A, B, C, D, E$.