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Introduction To Integration By Expansion Using Partial Fractions

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This section deals with the integration of general algebraic rational functions, of the form \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) , where \(f(x)\,\,{\rm{and \;}}g(x)\) are both polynomials. We already have seen some examples of this form. For example, we know how to integrate functions of the form \(\begin{align}\frac{1}{{Q(x)}}{\rm{ or }}\frac{{L(x)}}{{Q(x)}}{\rm{ or }}\frac{{P(x)}}{{Q(x)}}where\;L(x)\end{align}\)   where 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 where \(g(x)\) (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.

\(g(x) = {L_1}(x)\,{L_2}(x)\,...\,{L_n}(x)\) \(\text{(n linear factors)}\)
* \(g(x) = {L_1}(x)...L_r^k(x)...{L_n}(x)\)       \(\left( \begin{align}
  & n\,\,\text{linear factors; the }{{r}^{\text{th}}} \\ 
 & \text{factor is repeated }k\text{ times} \\ 
\end{align} \right)\)
\(g(x) = L_1^{{k_1}}(x)L_2^{{k_2}}(x)...L_n^{{k_n}}(x)\) \(\left( \begin{align}
  & n\,\,\text{linear factors, the }{{i}^{\text{th}}}\text{ factor } \\ 
 & \text{is repeated }{{k}_{\text{i}}}\text{ times} \\ 
\end{align} \right)\)
* \(g(x) = {L_1}(x){L_2}(x)...{L_n}(x)\,{Q_1}(x)\,{Q_2}(x)...{Q_m}(x)\)      \(\left( \begin{align}
  & n\,\,\text{linear factors and } \\ 
 & m\,\,\text{quadratic factors} \\ 
\end{align} \right)\)
*  \(g(x) = ...Q_r^k(x)...\)            \(\left( \begin{align}
  & \text{a particular quadratic factor } \\ 
 & \text{repeats more than once} \\ 
\end{align} \right)\) 
* A combination of any of the above.

                                                                                                                     

Suppose that the degree of \(g(x)\,\,{\text{is }}n\) and that of \(f(x)\,\,{\text{is }}m\) . If \(m \ge n,\) we can always divide \(f(x)\,\,{\text{by }}g(x)\) to obtain a quotient \(q(x)\) and a remainder \(r(x)\) whose degree would be less than n.

\[\frac{{f(x)}}{{g(x)}} = q(x) + \frac{{r(x)}}{{g(x)}}\qquad\qquad...{\rm{ }}\left( 1 \right)\]

If \(\begin{align}m < n,\,\,\frac{{f(x)}}{{g(x)}}\end{align}\) is termed a proper rational function.

The partial fraction 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 fractions.

From now one, we consider only proper rational functions. If  \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) is not proper, we make it proper \(\begin{align}\left( {\frac{{r(x)}}{{g(x)}}} \right)\end{align}\) by the procedure described in (1) above.

Let us consider a few examples.

Let \(g(x)\) be a product of non-repeated, linear factors:

\[g(x) = {L_1}(x){L_2}(x)...{L_n}(x)\]

Then, we can expand \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) in terms of partial fractions as

\[\begin{align}\frac{{f(x)}}{{g(x)}} = \frac{{{A_1}}}{{{L_1}(x)}} + \frac{{{A_2}}}{{{L_2}(x)}} + ... + \frac{{{A_n}}}{{{L_n}(x)}}\end{align}\]

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 fraction expansion of \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) :

\[\frac{{f(x)}}{{g(x)}} = \frac{{x + 1}}{{(x - 1)(x - 2)(x - 3)}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}}\]

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).

\[\begin{align}\;\;\;x = 1 \quad \,\,\, &\Rightarrow \quad 2 = A( - 1)( - 2) + B(0) + C(0)\\& \Rightarrow \quad A = 1 \\ \;x = 2 \quad \,\,\, &\Rightarrow \quad 3 = A(0) + B(1)( - 1) + C(0)\\ &\Rightarrow \quad B = - 3 \\ x = 3\, \quad \,\,\, &\Rightarrow \quad 4 = A(0) + B(0) + C(2)(1)\\& \Rightarrow \quad C = 2\end{align}\]
 

Thus, \(A = 1,\,\,B = - 3\,\,{\text{and }}C = 2\) .

We can therefore write \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) as a sum of partial fractions.

\[\frac{{f(x)}}{{g(x)}} = \frac{1}{{x - 1}} - \frac{3}{{x - 2}} + \frac{2}{{x - 3}}\]

Integrating \(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) is now a simple matter of integrating the partial fractions. This was our sole motive in writing such an expansion, so that integration could be carried out easily. In the example above:

\[\int {\frac{{f(x)}}{{g(x)}}} \,\,dx = \ln (x - 1) - 3\ln (x - 2) + 2\ln (x - 3) + C\]

Now, suppose that \(g(x)\) contains all linear factors, but a particular factor, say \({L_1}(x)\) , is repeated k times.

Thus,

\[g(x) = L_1^k(x)\,{L_2}(x)...{L_n}(x)\]

\(\begin{align}\frac{{f(x)}}{{g(x)}}\end{align}\) can now be expanded into partial fractions as follows:

\[\frac{{f(x)}}{{g(x)}} = \underbrace {\frac{{{A_1}}}{{{L_1}(x)}} + \frac{{{A_2}}}{{L_1^2(x)}} + \frac{{{A_3}}}{{L_1^3(x)}} + ...\frac{{{A_k}}}{{L_1^k(x)}}}_{k\,\,{\text{partial  fractions  corresponding  to }}{L_1}(x)} + \frac{{{B_2}}}{{{L_2}(x)}} + ... + \frac{{{B_n}}}{{{L_n}(x)}}\]

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.

\[\begin{align}\begin{array}{*{20}{c}}{\frac{1}{{{{(x - 1)}^2}(x - 2)}}}&{{\rm{can\; be\; expanded as\;}}}&{\frac{A}{{x - 1}} + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{x - 2}}}\\{\frac{1}{{{{(x - 1)}^3}(x - 2)(x - 3)}}}&{{\rm{can\; be \;expanded \;as\;}}}&{\frac{A}{{x - 1}} + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{{{(x - 1)}^3}}} + \frac{D}{{(x - 2)}} + \frac{C}{{(x - 3)}}}\\{\frac{1}{{{{(x - 1)}^2}{{(x + 5)}^3}}}}&{{\rm{can\; be \;expanded \;as\;}}}&{\frac{A}{{x - 1}} + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{(x + 5)}} + \frac{D}{{{{(x + 5)}^2}}} + \frac{E}{{{{(x + 5)}^3}}}}\end{array}\end{align}\]

Example - 36    Expand the following using partial fractions:

(a) \(\begin{align}\frac{{{x^2} - 3x - 4}}{{{x^3} - 6{x^2} + 11x - 6}}\end{align}\)

(b) \(\begin{align}\frac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}}\end{align}\)

Solution: (a) The denominator can be factorised:

\[{x^3} - 6{x^2} + 11x - 6 = (x - 1)(x - 2)(x - 3)\]

The partial fraction expansion is:

\[\frac{{(x + 1)(x - 4)}}{{(x - 1)(x - 2)(x - 3)}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}}\]

Cross-multiplying, we obtain

\[(x + 1)(x - 4) = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)\]

\[\begin{align}\begin{array}{*{20}{c}}{{\rm{Put }}x = 1}& \Rightarrow &{A = \frac{{(2) \times ( - 3)}}{{( - 1) \times ( - 2)}} = - 3}\\{{\rm{Put }}x = 2}& \Rightarrow &{B = \frac{{(3) \times ( - 2)}}{{(1) \times ( - 1)}} = 6\,\,\,\,\,\,\,}\\{{\rm{Put }}x = 3}& \Rightarrow &{C = \frac{{(4) \times ( - 1)}}{{(2) \times (1)}} = - 2\,\,\,\,}\end{array}\end{align}\]

Thus, the partial fraction expansion is

\[\frac{{ - 3}}{{x - 1}} + \frac{6}{{x - 2}} + \frac{{ - 2}}{{x - 3}}\]

(b) As discussed previously, the partial fraction expansion of this expression would be of the form

\[\frac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}} = \frac{A}{{x - 1}} + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{x - 2}} + \frac{D}{{x - 3}}\]

Cross-multiplying, we obtain

\[{x^2} + x + 1 = A(x - 1)(x - 2)(x - 3) + B(x - 2)(x - 3) + C{(x - 1)^2}(x - 3) + D{(x - 1)^2}(x - 2)\]

\[\begin{align}{}{{\text{Put }}\;x = 1}&\qquad \Rightarrow &{B = \frac{{1 + 1 + 1}}{{( - 1) \times ( - 2)}} = \frac{3}{2}}\\{{\text{Put }}\;x = 2}&\qquad \Rightarrow &{C = \frac{{4 + 2 + 1}}{{{1^2} \times - 1}} = - 7\,\,\,\,\,\,\,}\\{{\text{Put }}\;x = 3}&\qquad \Rightarrow &{D = \frac{{9 + 3 + 1}}{{{2^2} \times 1}} = \frac{{13}}{4}\,\,\,\,}\end{align}\]

To obtain A, we compare the coefficients of \({x^3}\) on both sides. Thus,

\[0\text{ }=A+C+D\]

\[\begin{align}& \Rightarrow\quad A = - (C + D) \\ &\qquad\quad\;= - \left( { - 7 + \frac{{13}}{4}} \right)\\&\qquad\quad\; = \frac{{15}}{4}\end{align}\]

The required partial fraction expansion is

\[\frac{{15/4}}{{x - 1}} + \frac{{3/2}}{{{{(x - 1)}^2}}} + \frac{{ - 7}}{{x - 2}} + \frac{{13/4}}{{x - 3}}\]

Download SOLVED Practice Questions of Introduction To Integration By Expansion Using Partial Fractions for FREE
Indefinite Integration
grade 11 | Questions Set 1
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Download SOLVED Practice Questions of Introduction To Integration By Expansion Using Partial Fractions for FREE
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
grade 11 | Answers Set 1
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration
grade 11 | Answers Set 2
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