Q19E
Question
In Problems 11–20, determine the partial fraction expansions for the given rational function.
\(\frac{1}{{(s - 3)\left( {{s^2} + 2s + 2} \right)}}\)
Step-by-Step Solution
VerifiedThe partial fraction expansions for the given rational function is \[\frac{1}{{\left( {s - 3} \right)\left( {{s^2} + 2s + 2} \right)}} = \frac{1}{{17}}\left[ {\frac{1}{{s - 3}} - \frac{{s + 1}}{{{{\left( {{s^2} + 1} \right)}^2} + 1}} - \frac{4}{{{{\left( {{s^2} + 1} \right)}^2} + 1}}} \right]\]
Any number which can be easily represented in the form of , such that \(p\) and \(q\) are integers and \(q \ne 0\) is known as a rational number. Similarly, we can define a rational function as the ratio of two polynomial functions \(P(x)\) and \(Q(x)\), where \(P\) and \(Q\) are polynomials in \(x\) and \(Q(x) \ne 0\). A rational function is known as proper if the degree of \(P(x)\) is less than the degree of \(Q(x)\); otherwise, it is known as an improper rational function. With the help of the long division process, we can reduce improper rational functions to proper rational functions. Therefore, if \(\frac{{{\rm{P}}({\rm{x}})}}{{{\rm{Q}}({\rm{x}})}}\) is improper, then it can be expressed as:
\(\frac{{P(x)}}{{Q(x)}} = A(x) + \frac{{R(x)}}{{Q(x)}}\)
Here, \(A(x)\) is a polynomial in \(x\) and \(\frac{{R(x)}}{{Q(x)}}\) is a proper rational function.
We know that the integration of a function \(f(x)\) is given by \(F(x)\) and it is represented by:
\(\int f (x)dx = F(x) + C\)
Here R.H.S. of the equation means integral of \(f(x)\) with respect to \(x\) and \(C\) is the constant of integration.
Write as sum of partial fractions
\(\frac{1}{{(s - 3)\left( {{s^2} + 2s + 2} \right)}} = \frac{A}{{s - 3}} + \frac{{Bs + C}}{{{s^2} + 2s + 2}}\)
Multiply both sides by the LCD \((s + 1)(s - 2)(s - 1)\)
\(1 = A\left( {{s^2} + 2s + 2} \right) + (Bs + C)(s - 3)\)
Simplify
\(1 = A{s^2} + 2As + 2A + B{s^2} - 3Bs + Cs - 3C\)
Group like terms and factor
\(\begin{array}{l}1 = \left( {A{s^2} + B{s^2}} \right) + (2As - 3Bs + Cs) + (2A - 3C)\\1 = (A + B){s^2} + (2A - 3B + C)s + (2A - 3C)\end{array}\)
Compare coefficients
\(1 = \underbrace {(A + B)}_0{s^2} + \underbrace {(2A - 3B + C)}_0s + \underbrace {(2A - 3C)}_1\)
Therefore,
\(\left\{ {\begin{array}{*{20}{c}}{A + B = 0}\\{2A - 3B + C = 0}\\{2A - 3C = 1}\end{array}} \right.\)
Find the constants, solve the system, you get:
\(A = \frac{1}{{17}},B = - \frac{1}{{17}},C = - \frac{5}{{17}}\)
Substitute the constants:
\(\begin{array}{c}\frac{1}{{(s - 3)\left( {{s^2} + 2s + 2} \right)}} = \frac{1}{{17(s - 3)}} + \frac{{ - s - 5}}{{17\left( {{s^2} + 2s + 2} \right)}}\\\frac{1}{{(s - 3)\left( {{s^2} + 2s + 2} \right)}} = \frac{1}{{17}}\left[ {\frac{1}{{s - 3}} - \frac{{s + 5}}{{\left( {{s^2} + 2s + 2} \right)}}} \right]\\ = \frac{1}{{17}}\left[ {\frac{1}{{s - 3}} - \frac{{s + 4 + 1}}{{{{\left( {{s^2} + 1} \right)}^2} + 1}}} \right]\\ = \frac{1}{{17}}\left[ {\frac{1}{{s - 3}} - \frac{{s + 1}}{{{{\left( {{s^2} + 1} \right)}^2} + 1}} - \frac{4}{{{{\left( {{s^2} + 1} \right)}^2} + 1}}} \right]\end{array}\)
The partial fraction expansions for the given rational function is \[\frac{1}{{\left( {s - 3} \right)\left( {{s^2} + 2s + 2} \right)}} = \frac{1}{{17}}\left[ {\frac{1}{{s - 3}} - \frac{{s + 1}}{{{{\left( {{s^2} + 1} \right)}^2} + 1}} - \frac{4}{{{{\left( {{s^2} + 1} \right)}^2} + 1}}} \right]\]