Repeated Linear Factors

A repeated linear factor $(x-a)^n$ in the denominator cannot be given a single constant numerator $A/(x-a)^n$ -- the correct decomposition needs an entire descending chain of powers:

After clearing denominators, substituting $x=a$ only ever recovers the last constant $A_n$ directly, since every other term on the right still carries a positive power of $(x-a)$ and vanishes at $x=a$. The remaining constants can be found by comparing coefficients of the surviving powers of $x$ -- workable, but often tedious for a high power $n$.

A cleaner route, used throughout this section, is the shift trick: put $t=x-a$ (so $x=t+a$), rewrite the numerator of the original fraction as a polynomial in $t$ by direct substitution, and then divide every term of that polynomial by $t^n$. Each power of $t$ in the numerator then lines up automatically with the matching negative power of $t$ -- exactly the matching partial-fraction term. For instance, resolving $\dfrac{x^2+1}{(x-1)^3}$: put $t=x-1$, so the numerator becomes $(t+1)^2+1=t^2+2t+2$, and

$\frac{t^2+2t+2}{t^3}=\frac1t+\frac2{t^2}+\frac2{t^3}=\frac1{x-1}+\frac2{(x-1)^2}+\frac2{(x-1)^3},$ …