How to find the partial fraction when the denominator contain repeated root?

Question

I do not know how to find the partial fraction of the following form since the denominator contains a multiplicity of the root with an integer degree $n \geq1$

$$\frac{1}{{\left( {1 + x} \right){{\left( {x + \alpha } \right)}^n}}} = \frac{A}{{\left( {1 + x} \right)}} + \frac{B}{{{{\left( {x + \alpha } \right)}^n}}} +\cdots$$

Answer 1

@Thai-Hoc Vu I believe the answer would be

$\frac{1}{{{{\left( {\alpha - 1} \right)}^n}\left( {x + 1} \right)}} - \sum\limits_{k = 1}^n {\frac{1}{{{{\left( {\alpha - 1} \right)}^{n - k + 1}}{{\left( {\alpha + x} \right)}^k}}}}$

You can verify it on wolfram alpha Wolfram Alpha

Answer 2

The form to use is $${A\over1+x}+{B_1\over x+a}+{B_2\over(x+a)^2}+\cdots+{B_n\over(x+a)^n}$$ [assuming $a\ne1$]

Answer 3

Sometimes, for example for use in integrals, could be handy the decomposition for $n>1$ $$ \frac{P_{n-1}(x)}{(x-x_0)^n}=\frac{A}{x-x_0}+\frac{d}{dx}\left[\frac{Q_{n-2}(x)}{(x-x_0)^{n-1}}\right] $$ where $A$ and the coefficients of the $n-2$ degree polynomial $Q_{n-2}(x)$ have to be determined by polynomial identity.

Answer 4

The general method, when you have a proper rational function (i.e. the degree of the numerator is less than the degree of the denominator) with a pole of order $n$, consists in performing the division of the numerator by the other factor by increasing powers, up to order $n$.

Here is how it goes here: first it simplifies the computations if the pole of order $n$ is $0$. So we begin by the substitution $t=x+\alpha\iff x=t-\alpha$, which yields the fraction $$\frac1{(t-\alpha+1)\,t^n}.$$ To illustrate the procedure, I'll suppose $n=3$. This is the result of division by increasing powers of $t$ up to order $3$: $$\begin{array}{rcl} \color{red}{\frac1{1-\alpha}-\frac t{(1-\alpha)^2}+\frac{t^2}{(1-\alpha)^3}}\\ \hline 1-\alpha+t &\Bigl(\begin{alignedat}[t]{5} 1 \\ -1&&{}-\frac{t}{1-\alpha}\\ \hline &&-\frac{t}{1-\alpha}\\&&\frac{t}{1-\alpha}&&{}+\frac{t^2}{(1-\alpha)^2} \\\hline &&&&{}\frac{t^2}{(1-\alpha)^2} \\ &&&&{}-\frac{t^2}{(1-\alpha)^2} &&{}-\frac{t^3}{(1-\alpha)^3} \\\hline &&&&&&\color{red}{{}- \frac{t^3}{(1-\alpha)^3} } \end{alignedat} \end{array}$$

There for from the equality $$1=(1-\alpha+t)\biggl(\frac1{1-\alpha}-\frac t{(1-\alpha)^2}+\frac{t^2}{(1-\alpha)^3}\biggr)-\frac{t^3}{(1-\alpha)^3},$$ we deduce that $$\frac 1{(1-\alpha+t)\,t^3}=\frac1{(1-\alpha)t^3}-\frac 1{(1-\alpha)^2\,t^2}+\frac{1}{(1-\alpha)^3\,t}-\frac{t^3}{(1-\alpha)^3(1-\alpha+t)}$$ Can you proceed?