Our inductive hypothesis here is as follows: \[ \frac{x}{(x+a)^k(x+b)}\equiv\frac{F}{x+b}+\frac{E_1}{x+a}+\frac{E_2}{(x+a)^2}+\dots+\frac{E_n}{(x+a)^k} \] Our goal is to prove RHS numerator constancy for power of \((k+1)\). \[ \frac{x}{(x+a)^{k+1}(x+b)}\equiv\frac{1}{x+a}\times\frac{x}{(x+a)^k(x+b)} \] \[ \equiv\frac{1}{x+a}\times\left(\frac{F}{x+b}+\frac{E_1}{x+a}+\frac{E_2}{(x+a)^2}+\dots+\frac{E_n}{(x+a)^k}\right) \] \[ \equiv\frac{F}{(x+a)(x+b)}+\frac{E_1}{(x+a)^2}+\frac{E_2}{(x+a)^3}+\dots+\frac{E_n}{(x+a)^{k+1}} \] The first term here is exactly the same as our base case (for one in the numerator), so we can substitute accordingly: \[ \equiv\frac{F_1}{(x+a)}+\frac{F_2}{(x+b)}+\frac{E_1}{(x+a)^2}+\frac{E_2}{(x+a)^3}+\dots+\frac{E_n}{(x+a)^{k+1}} \] \[ \textrm{therefore} \quad \frac{1}{(x+a)^n(x+b)}\equiv\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_n}{(x+a)^n} \] This concludes the first half of this section's proof.
With \(n\) out of the way, we turn our attention to \(m\). Thus, our aim is now the overall aim of this section, constant \(A,B\)s where \[ \frac{x}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \textrm{.} \] We are again going to use induction, and we see that we have already proved our base case in the preceding half of this section, \[ \frac{x}{(x+a)^n(x+b)}\equiv\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_n}{(x+a)^n} \textrm{.} \] Moving on to the inductive step, our assumption here is \[ \frac{x}{(x+a)^n(x+b)^k}\equiv\sum_{r=1}^n\frac{E_r}{(x+a)^r}+\sum_{r=1}^k\frac{F_r}{(x+b)^r} \textrm{.} \] We will now show that when the power is \((k+1)\), the numerators are still constant: \[ \frac{x}{(x+a)^n(x+b)^{k+1}}\equiv\frac{1}{x+b}\times\frac{x}{(x+a)^n(x+b)^k} \] \[ \equiv\sum_{r=1}^n\frac{E_r}{(x+a)^r(x+b)}+\sum_{r=1}^k\frac{F_r}{(x+b)^{r+1}} \] The contents of the first sum is identical to our base case, so we will substitute that in: \[ \equiv\sum_{r=1}^nE_r\left(\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_r}{(x+a)^r}\right)+\sum_{r=1}^k\frac{F_r}{(x+b)^{r+1}} \] At this point it is clear the only terms left remaining after expansion will be those compliant with our proposition, so we may proceed to: \[ \frac{x}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \] This proves our aim.