I need to make a partial fraction decomposition of the following fraction :
$$ \frac{1}{(x-a)^2(x-b)^2(x-c)^2(x-d)^2(x-e)} $$
The problem is that Wolfram doesn't give any answer : http://wolfr.am/49l6Tjzy
Could someone help me please ?
More precisly I am trying to find a recursive formula for this kind of fractions :
$$ \frac{1}{(x-a)^2(x-b)^2(x-c)^2(x-d)^2(x-e)^2(x-f)} $$
Thank you for your answers.
Let: $$ h(a_1,\ldots,a_n;x)=\prod_{j=1}^{n}(x-a_j)^{-1}.\tag{1}$$ By the residue theorem: $$ h(a_1,\ldots,a_n; x) = \sum_{j=1}^{n}\frac{\operatorname{Res}(h,x=a_j)}{x-a_j}=\sum_{j=1}^{n}\frac{1}{(x-a_j)\prod_{k\neq j}(a_k-a_j)}\tag{2}$$ and now we just need to multiply $h(a_1,\ldots,a_{n};x)$ and $h(a_1,\ldots,a_n,a_{n+1};x)$: $$h(a_1,\ldots,a_{n};x)\cdot h(a_1,\ldots,a_n,a_{n+1};x)=\textstyle{\left(\sum_{j=1}^{n}\frac{1}{(x-a_j)\prod_{k\neq j}(a_k-a_j)}\right)\cdot\left(\sum_{j=1}^{n}\frac{1}{(x-a_j)\prod_{k\neq j}(a_k-a_j)}+\frac{1}{(x-a_{n+1})\prod_{k=1}^{n}(a_j-a_{n+1})}\right)}\tag{3}$$ and by setting $A_j=\prod_{\substack{k=1\\k\neq j}}^{n}(a_k-a_j)$ and $A_{n+1}=\prod_{k=1}^{n}(a_k-a_{n+1})$ we just need to exploit: $$\frac{1}{(x-a_{j_1})A_{j_1}}\cdot \frac{1}{(x-a_{j_2})A_{j_2}}=\frac{1}{(a_{j_1}-a_{j_2})A_{j_1}A_{j_2}}\left(\frac{1}{x-a_{j_1}}-\frac{1}{x-a_{j_2}}\right)\tag{4}$$ to find the full partial fraction decomposition of $(3)$.