Question

if \(a \ne 0\) and n is a positive integer, find the partial fraction decomposition of \(f(x) = \frac{1}{{{x^n}(x - a)}}\)

Short answer

The partial fraction decomposition of the given function can be given as:

\(\frac{1}{{{x^n}(x - a)}} = \frac{1}{{{a^n}(x - a)}} - \frac{1}{{{a^n}x}} - \frac{1}{{{a^{n - 1}}{x^2}}}....... - \frac{1}{{a{x^n}}}\)

Step-by-step solution

Find the partial fraction decomposition

\(f(x) = \frac{1}{{{x^n}(x - a)}}\)

We expand it like following

\(\begin{aligned}{l}\frac{1}{{{x^n}(x - a)}} = \frac{{{c_1}}}{{(x - a)}} + \frac{{{c_2}}}{x} + \frac{{{c_3}}}{{{x^2}}}....... + \frac{{{c_n}}}{{{x^n}}}\,\,\,\,\,\,\,\,\,\,\,\,....(1)\\1 = {c_1}{x^n} + (x - a){c_2}{x^{n - 1}} + (x - a){c_3}{x^{n - 2}} + ......... + {c_n}(x - a)\\0 = {c_1} + {c_2}\\0 = - a{c_2} - {c_3}\\.\\.\\.\\.\\0 = - a{c_{n - 1}} + {c_n}\\1 = - a{c_n}\\{c_n} = \frac{{ - 1}}{a}\\{c_{n - 1}} = \frac{{ - 1}}{{{a^2}}}\end{aligned}\)

Substitute all the values of c in the equation (1)

So we get,

\(\begin{aligned}{l}\frac{1}{{{x^n}(x - a)}} = \frac{{{c_1}}}{{(x - a)}} + \frac{{{c_2}}}{x} + \frac{{{c_3}}}{{{x^2}}}....... + \frac{{{c_n}}}{{{x^n}}}\,\,\,\,\,\\\frac{1}{{{x^n}(x - a)}} = \frac{1}{{{a^n}(x - a)}} - \frac{1}{{{a^n}x}} - \frac{1}{{{a^{n - 1}}{x^2}}}....... - \frac{1}{{a{x^n}}}\end{aligned}\)

Hence, the partial fraction decomposition of the given function \(f(x) = \frac{1}{{{x^n}(x - a)}}\) can be given as:

\(\frac{1}{{{x^n}(x - a)}} = \frac{1}{{{a^n}(x - a)}} - \frac{1}{{{a^n}x}} - \frac{1}{{{a^{n - 1}}{x^2}}}....... - \frac{1}{{a{x^n}}}\)