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## Monday, 4 August 2014

### CHAPTER 10 - Expansion Using Partial Fractions

This section deals with the integration of general algebraic rational functions, of the form $\dfrac{{f(x)}}{{g(x)}}$, where $f(x)\,$ and $g(x)$ are both polynomials. We already have seen some examples of this form. For example, we know how to integrate functions of the form $\dfrac{1}{{Q(x)}}$ or $\dfrac{{L(x)}}{{Q(x)}}$ or $\dfrac{{P(x)}}{{Q(x)}}$ where $L(x)$ is a linear factor, $Q(x)$ is a quadratic factor and $P(x)$ is a polynomial of degree $n \ge 2$. We intend to generalise that previous discussion in this section.
We are assuming the scenario $g(x)$ where (the denominator) is decomposible into linear or quadratic factors. These are the only cases relevant to us right now. Any linear or quadratic factor in $g(x)$ might also occur repeatedly.
Thus, $g(x)$ could be of the following general forms.
•  $g(x) = {L_1}(x)\,{L_2}(x)\,\ldots\,{L_n}(x)$ ($n$ linear factors)
•  $g(x) = {L_1}(x)\ldots L_r^k(x).\ldots {L_n}(x)$ ($n$ linear factors; the ${r^{th}}$factor is repeated $k$ times)
•  $g(x) = L_1^{{k_1}}(x)L_2^{{k_2}}(x)\ldots L_n^{{k_n}}(x)$ ($n$ linear factors, the ${i^{th}}$ factor is repated ${k_i}$ times)
•  $g(x) = {L_1}(x){L_2}(x)\ldots {L_n}(x)\,{Q_1}(x)\,{Q_2}(x)\ldots {Q_m}(x)$ ($n$ linear factors and $m$quadratic factors)
•  $g(x) = \ldots Q_r^k(x)\ldots$ (a particular quadratic factors repeat more than once)
• A combination of any of the above.
Suppose that the degree of $g(x)\,$ is $n$ and that of $f(x)$ is $m$. If $m \ge n,$ we can always divide $f(x)$ by $g(x)$ to obtain a quotient $q(x)$ and a remainder $r(x)$whose degree would be less than $n$.
 $\dfrac{{f(x)}}{{g(x)}} = q(x) + \dfrac{{r(x)}}{{g(x)}}$ $\ldots(1)$
If $m < n,\,\,\dfrac{{f(x)}}{{g(x)}}$ is termed a proper rational function.
The partial dfraction expansion technique says that a proper rational function can be expressed as a sum of simpler rational functions each possessing one of the factors of $g(x)$. The simpler rational functions are called partial dfractions.
From now one, we consider only proper rational functions. If $\dfrac{{f(x)}}{{g(x)}}$ is not proper, we make it proper $\left( {\dfrac{{r(x)}}{{g(x)}}} \right)$ by the procedure described in ($1$) above.
Let us consider a few examples.
Let $g(x)$ be a product of non-repeated, linear factors:
 $g(x) = {L_1}(x){L_2}(x)\ldots {L_n}(x)$
Then, we can expand $\dfrac{{f(x)}}{{g(x)}}$ in terms of partial dfractions as
 $\dfrac{{f(x)}}{{g(x)}} = \dfrac{{{A_1}}}{{{L_1}(x)}} + \dfrac{{{A_2}}}{{{L_2}(x)}} + \ldots + \dfrac{{{A_n}}}{{{L_n}(x)}}$
where the ${A_i}^\prime s$ are all constants that need to be determined
Suppose $f(x) = x + 1$ and $g(x) = (x - 1)(x - 2)(x - 3)$. Let us write down the partial dfraction expansion of $\dfrac{{f(x)}}{{g(x)}}$:
 $\dfrac{{f(x)}}{{g(x)}} = \dfrac{{x + 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{x - 2}} + \dfrac{C}{{x - 3}}$
We need to determine $A$$B$ and $C$. Cross multiplying in the expression above, we obtain:
 $(x + 1) = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$
$A$$B$$C$ can now be determined by comparing coefficients on both sides. More simply since this relation that we’ve obtained should held true for all $x$, we substitute those values of $x$ that would straight way give us the required values of $A$$B$ and $C$. These values are obviously the roots of $g(x)$.
 $x = 1 \,\,\, \Rightarrow 2 = A( - 1)( - 2) + B(0) + C(0)$ $\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 1$ $x = 2 \,\,\, \Rightarrow \,\,\,\, 3 = A(0) + B(1)( - 1) + C(0)$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\, B = - 3$ $x = 3\, \,\,\, \Rightarrow \,\,\,\, 4 = A(0) + B(0) + C(2)(1)$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\, C = 2$
Thus, $A = 1,\,\,B = - 3$ and $C = 2$.
We can therefore write $\dfrac{{f(x)}}{{g(x)}}$ as a sum of partial fractions.
 $\dfrac{{f(x)}}{{g(x)}} = \dfrac{1}{{x - 1}} - \dfrac{3}{{x - 2}} + \dfrac{2}{{x - 3}}$
Integrating $\dfrac{{f(x)}}{{g(x)}}$ is now a simple matter of integrating the partial dfractions. This was our sole motive in writing such an expansion, so that integration could be carried out easily. In the example above:
 $\int {\dfrac{{f(x)}}{{g(x)}}} \,\,dx = \ln (x - 1) - 3\ln (x - 2) + 2\ln (x - 3) + C$
Now, suppose that $g(x)$ contains all linear factors, but a particular factor, say ${L_1}(x)$, is repeated $k$ times.
Thus,
 $g(x) = L_1^k(x)\,{L_2}(x)\ldots {L_n}(x)$
$\dfrac{{f(x)}}{{g(x)}}$ can now be expanded into partial dfractions as follows:
 $\dfrac{{f(x)}}{{g(x)}} = \underbrace {\dfrac{{{A_1}}}{{{L_1}(x)}} + \dfrac{{{A_2}}}{{L_1^2(x)}} + \dfrac{{{A_3}}}{{L_1^3(x)}} +\ldots \dfrac{{{A_k}}}{{L_1^k(x)}}}_{Comment} + \dfrac{{{B_2}}}{{{L_2}(x)}} + \ldots + \dfrac{{{B_n}}}{{{L_n}(x)}}$
Comment: $k$ partial dfractions corresponding to ${L_1}(x)$
This means that we will have $k$ terms corresponding to ${L_1}(x)$. The rest of the linear factors will have single corresponding terms in the expansion. Here are some examples.
 ${\dfrac{1}{{{{(x - 1)}^2}(x - 2)}}}$ can be expanded as ${\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{x - 2}}}$ ${\dfrac{1}{{{{(x - 1)}^3}(x - 2)(x - 3)}}}$ can be expanded as ${\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{{{(x - 1)}^3}}} + \dfrac{D}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}}$ ${\dfrac{1}{{{{(x - 1)}^2}{{(x + 5)}^3}}}}$ can be expanded as ${\dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 5)}} + \dfrac{D}{{{{(x + 5)}^2}}} + \dfrac{E}{{{{(x + 5)}^3}}}}$