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

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 $C',D',E'$.
