Integration by Partial Fractions: Numerator Should Be One Degree Lower than Denominator

Question

At 1:36:13 (on https://www.youtube.com/watch?v=KJGp0pyPoVo&list=PLDesaqWTN6EQ2J4vgsN1HyBeRADEh4Cw-&index=11), the professor says:

Whatever factor you have, your numerator should be one degree lower than that. So for linears, one degree lower than a linear is a constant. One degree lower than a quadratic is a linear.

However, this doesn't seem to hold true in some of the worked examples in Stewart's Early Transcendentals (8th Edition) where you have:

$$\frac{x^3-x+1}{x^2(x-1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}$$

Another example shows:

$$\frac{4x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$$

With D and E in the first equation, and B in the second equation being constants, don't these examples disprove the statement above?

Answer 1

It's just an alternative way of writing it, since

$$\frac{A}{x} + \frac{B}{x^2} = \frac{Ax + B}{x^2}$$

and the polynomial $Ax+B$ is one degree lower than $x^2$.

---

If the polynomial isn't $x^n$, then the constants change a little, since

$$\frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3} = \frac{C(x-1)^2 + D(x-1) + E}{(x-1)^3} = \frac{C'x^2 + D'x + E'}{(x-1)^3}$$

for some constants $C', D', E'$, so it's sort of up to you. You can either go calculate $C,D,E$ or $C',D',E'$.

Answer 2

The exact formulation is that the degree of the numerator should be lower than the degree of the irreducible factor in the denominator, not than the degree of the denominator itself.

And it is not necessarily lower by $1$: if the degree of the irreducible factor in the denominator is $d$, the degree of the numerator is at most $d-1$.