What is the partial fraction decomposition of $f(x) = \dfrac{1}{x^n(1-x)}$

You may use that $1-x^n = (1-x)(x^{n-1} + x^{n-2}+\cdots +x+1)$

Can someone help me with this one? This is what I have so far:

$f(x) = \dfrac{1}{x^n(1-x)} = \dfrac{A_0}{x}+\dfrac{A_1}{x^2}+\dfrac{A_2}{x^3}+\cdots +\dfrac{A_{n}}{x^n}+\dfrac{A_{n+1}}{1-x}$.

Multiplying through $x^n(1-x)$, and rearranging, I get something similar to the hint that was given:

$1-A_{n+1}x^n = (1-x)( A_0x^{n-1} + A_1x^{n-2}+ A_2x^{n-3}+\cdots+A_{n-1}x+A_n)$.

Now I am not sure how to proceed, it's only a 2 point question, so it doesn't seem like it should be difficult, but how do I solve for the constants?

I greatly appreciate any help, point in the right direction!

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So, in your last line, if you let $A_i = 1$ for all $i$, you are done by line 1. –  martini Sep 26 '12 at 12:04

$$\tag{1} \frac{1}{x^n(1-x)}=\sum_{k=1}^n\frac{A_k}{x^k}+\frac{A_{n+1}}{1-x}.$$ Multiplying both sides of (1) by $1-x$ and setting $x=1$, we have $A_{n+1}=1$. So now we have for $x \ne 1$: \begin{eqnarray} \sum_{k=1}^n\frac{A_k}{x^k}&=&\frac{1}{x^n(1-x)}-\frac{1}{1-x}=\frac{1}{1-x}\left(\frac{1}{x^n}-1\right)=\frac{1}{1-x}\left(\frac{1}{x}-1\right)\sum_{k=0}^{n-1}\frac{1}{x^k}\cr &=&\frac{1}{x}\sum_{k=0}^{n-1}\frac{1}{x^k}=\sum_{k=1}^n\frac{1}{x^k}, \end{eqnarray} i.e. $A_k=1$ for every $1 \le k \le n+1$, and $$\frac{1}{x^n(1-x)}=\sum_{k=1}^n\frac{1}{x^k}+\frac{1}{1-x}.$$

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Thanks for your help Mercy! If I were to then integrate this - how would I go about integrating the first part, with the summation? –  JackReacher Sep 26 '12 at 22:02
You just integrate each term, as long as it makes sense! –  Mercy Sep 26 '12 at 22:48
Note that $$\frac{1}{x^n(1-x)} = \frac{1}{x^n} + \frac{1}{x^{n-1}(1-x)}$$ Now use induction.
If we expand right hand side of your last expression, we get $$A_n+(A_{n-1}-A_{n})x+(A_{n-2}-A_{n-1})x^2+\cdots+(A_{1}-A_{0})x^{n-1}-A_0x^{n}.$$ Therefore, we get $$1-A_{n+1}x^n=A_n+(A_{n-1}-A_{n})x+(A_{n-2}-A_{n-1})x^2+\cdots+(A_{1}-A_{0})x^{n-1}-A_0x^{n}.$$ Comparing the coefficients of the like terms, we obtain $$1=A_n, A_{n-1}-A_{n}=0, A_{n-1}-A_{n-2}+0,..., A_{1}-A_{0}=0,\mbox{ and }A_{n+1}=A_0.$$ This implies that $$A_i=1\mbox{ for all }i=0,1,...,n+1.$$