Examples On Integration By Expansion Using Partial Fractions Set-1

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Example - 37

Evaluate \(\begin{align}\int {\frac{{{x^2} - 8x + 7}}{{{{(x - 5)}^2}{{(x + 2)}^2}}}\,\,dx}\end{align} \)

Solution: We first express this expression as a sum of partial fractions:

\[\frac{{(x - 1)(x - 7)}}{{{{(x - 5)}^2}{{(x + 2)}^2}}} = \frac{A}{{x - 5}} + \frac{B}{{{{(x - 5)}^2}}} + \frac{C}{{x + 2}} + \frac{D}{{{{(x + 2)}^2}}}\]

Cross-multiplying, we obtain

\[(x - 1)(x - 7) = A(x - 5){(x + 2)^2} + B{(x + 2)^2} + C{(x - 5)^2}(x + 2) + D{(x - 5)^2}\]

\[\begin{align}{}&{{\text{Put }}x = 5\,\,\,}& \Rightarrow \quad &{B = \frac{{4 \times - 2}}{{{7^2}}} = \frac{{ - 8}}{{49}}\,}\\&{{\text{Put }}x = - 2}& \Rightarrow\quad &{D = \frac{{ - 3 \times - 9}}{{{7^2}}} = \frac{{27}}{{49}}}\end{align}\]

To obtain A and C, we can now compare coefficients on both sides.

Comparing the coefficients of \({x^3}\) on both sides, we obtain

\[\begin{align}& \qquad\quad 0 = A + C\\ &\Rightarrow\quad\; A = - C\qquad \qquad...\left( 1 \right)\end{align}\]

Comparing the constant terms on both sides, we obtain

\[7 = - 20A + 4B + 50C + 25D\qquad \qquad ...\left( 2 \right)\]

Using (1) in (2), we obtain

\[\begin{align} &\qquad\qquad 7 = 20C + 4B + 50C + 25D\\ &\qquad\quad\quad\;\;= 70C - \frac{{32}}{{49}} + \frac{{675}}{{49}}\\& \Rightarrow\quad\; 70C = \frac{{ - 400}}{{49}}\\& \Rightarrow\quad\quad\; C = \frac{{ - 40}}{{343}}\\& \Rightarrow\quad\quad\; A = \frac{{40}}{{343}} \end{align}\]

Thus, the required integral is

\[I = \frac{{40}}{{343}}\ln \left| {x - 5} \right| + \frac{8}{{49(x - 5)}} - \frac{{40}}{{343}}\ln \left| {x + 2} \right| - \frac{{27}}{{49(x + 2)}} + C\]

We now come to the case when \(g(x)\) has non-repeating quadratic factors. Let \({Q_1}(x) = a{x^2} + bx + c\) be such a factor.

Our partial fraction expansion technique says that corresponding to \({Q_1}(x)\), there will be a term of the form \(\begin{align}\frac{{{L_1}(x)}}{{{Q_1}(x)}}\end{align}\) where \({L_1}(x)\) is a linear factor, i.e.

\[\frac{{f(x)}}{{g(x)}} = ...\underbrace { + \frac{{px + q}}{{a{x^2} + bx + c}}}_{{\text{factor corresponding to }}{Q_1}(x)} + ...\]

In the case that a quadratic factor, say \({Q_1}(x)\) , is repeating, for instance k times, there will be k corresponding partial fractions for \({Q_1}(x)\) .

\[\frac{{f(x)}}{{g(x)}} = ...\underbrace { + \frac{{{p_1}x + {q_1}}}{{a{x^2} + bx + c}} + \frac{{{p_2}x + {q_2}}}{{{{(a{x^2} + bx + c)}^2}}} + ...\frac{{{p_k}x + {q_k}}}{{{{(a{x^2} + bx + c)}^k}}}}_{k\,\,{\text{factors corresponding to }}{Q_1}(x)} + ...\]

Here are a few examples:

\[\begin{align}{*\;\;\;\;\;\;\;\frac{1}{{(x + 1)({x^2} + 2x + 3)}}}&\quad{{\text{can be expanded  as}}}\quad {\frac{A}{{x + 1}} + \frac{{Bx + C}}{{({x^2} + 2x + 3)}}}\\{*\;\;\;\;\;\;\;\frac{1}{{(x + 1){{({x^2} + 2x + 3)}^2}}}}&\quad{{\text{can be expanded as}}}\quad {\frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 3}} + \frac{{Dx + E}}{{{{({x^2} + 2x + 3)}^2}}}}\end{align}\] etc.

Example – 38

Evaluate the following integrals:

(a) \(\begin{align}\int {\frac{1}{{x\left( {1 + x + {x^2} + {x^3}} \right)}}} \,\,dx\end{align}\)

(b) \(\begin{align}\int {\frac{{3{x^2} + x + 3}}{{{{\left( {x - 1} \right)}^3}\left( {{x^2} + 1} \right)}}} \,\,dx\end{align}\)

Solution: (a)                                                                                  \(\begin{align} I = \int {\frac{1}{{x\left( {1 + x + {x^2}\left( {1 + x} \right)} \right)}}} \,\,dx\end{align}\)

\[ = \int {\frac{1}{{x\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}} \,\,dx\]

We find out the partial fraction expansion of this expression:

\[\frac{1}{{x\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}\, = \,\frac{A}{x} + \frac{B}{{1 + x}} + \frac{{Cx + D}}{{1 + {x^2}}}\]

Cross-multiplying, we obtain

\[1 = A\left( {1 + x} \right)\left( {1 + {x^2}} \right) + Bx\left( {1 + {x^2}} \right) + \left( {Cx + D} \right)\left( x \right)\left( {1 + x} \right)\]

\( Put \; x = 0 \quad\qquad \Rightarrow \qquad A = 1\)

\( Put \;  x = - 0 \qquad \Rightarrow \qquad B = - \frac{1}{2}\)

Compare the coeffs of \({x^3}\) \( \Rightarrow 0 = A + B + C\)

\[ \Rightarrow \qquad C = - \frac{1}{2}\]

Compare the coeffs of \({x^2}\) \( \Rightarrow 0 = A + C + D\)

\[ \Rightarrow \qquad D = - \frac{1}{2}\]

The partial fraction expansion is

\[\frac{1}{x} + \frac{{\left( { - 1/2} \right)}}{{1 + x}} + \frac{{\left( { - 1/2} \right)x + \left( { - 1/2} \right)}}{{1 + {x^2}}}\]

The integral is therefore:

\[\qquad\qquad I = \int {\frac{1}{x}dx - \frac{1}{2}\int {\frac{1}{{1 + x}}dx - \frac{1}{2}\int {\frac{x}{{1 + {x^2}}}dx - \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} \,dx} } } \]

\[ = \ln \left| x \right| - \frac{1}{2}\ln \left| {1 + x} \right| - \frac{1}{4}\ln \left( {1 + {x^2}} \right) - \frac{1}{2}{\tan ^{ - 1}}x + C\]

(b) We again find out the partial fraction expansion of the given expression:

\[\frac{{3{x^2} + x + 3}}{{{{\left( {x - 1} \right)}^3}\left( {{x^2} + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{{{\left( {x - 1} \right)}^3}}} + \frac{{Dx + E}}{{{x^2} + 1}}\]

We cross multiply to obtain

\[3{x^2} + x + 3 = A{\left( {x - 1} \right)^2}\left( {{x^2} + 1} \right) + B\left( {x - 1} \right)\left( {{x^2} + 1} \right) + C\left( {{x^2} + 1} \right) + \left( {Dx + E} \right){\left( {x - 1} \right)^3}\]

\[\begin{align}
 & Put\,\,\,x=1\qquad \qquad \quad \,\;\;\qquad \Rightarrow \qquad C=\frac{7}{2}\qquad\qquad \qquad \qquad \qquad \qquad \quad\ ...\text{ }\left( 1 \right) \\ 
 & \text{Compare the coeff  of  }\;{{x}^{4}}\qquad \Rightarrow \qquad 0=A+D\qquad\qquad\qquad\qquad\qquad\quad \;\;\;...\left( 2 \right) \\ 
 & \text{Compare the coeff  of  }\;{{x}^{3}}\qquad \Rightarrow \qquad 0=-2A+B-3D+E\qquad\qquad\qquad ...\left( 3 \right) \\ 
 & \text{Compare the coeff  of  }\;{{x}^{2}}\qquad \Rightarrow \qquad 3=2A-B+C+3D-3E\qquad\qquad\; ...\left( 4 \right) \\ 
 & \text{Compare the coeff of    }\; x\qquad  \;\;\Rightarrow  \qquad 1=-2A+B-D+3E\qquad\qquad\qquad ...\left( 5 \right) \\ 
\end{align}\]

Adding (4) and (5), we obtain

\[\begin{align} &\qquad\;\;\; 4 = C + 2D\\ &\Rightarrow\quad D = \frac{1}{4}\qquad\qquad\left( {from{\text{ }}\left( 1 \right)} \right)\\& \Rightarrow \quad A = \frac{{ - 1}}{4}\qquad\quad\;\left( {from{\text{ }}\left( 2 \right)} \right)\end{align}\]

Adding (3) and (4), we obtain

\[ 3 = C - 2E \\ \qquad\qquad \Rightarrow \qquad E = \frac{1}{4}\qquad \left( {again,{\text{ }}using{\text{ }}\left( 1 \right)} \right)\]

Finally, from (5), B = 0

The partial fraction expansion is therefore,

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

The integral is

\[\begin{align}& I = - \frac{1}{4}\int {\frac{1}{{x - 1}}} \,\,dx + \frac{7}{2}\int {\frac{1}{{{{\left( {x - 1} \right)}^3}}}} \,dx + \frac{1}{4}\int {\frac{x}{{{x^2} + 1}}dx + \frac{1}{4}} \int {\frac{1}{{{x^2} + 1}}} \,dx \\ & \;\;= - \frac{1}{2}\ln \left| {x - 1} \right| - \frac{7}{{4{{\left( {x - 1} \right)}^2}}} + \frac{1}{8}\ln \left( {{x^2} + 1} \right) + \frac{1}{4}{\tan ^{ - 1}}x + C\\&\;\; = \frac{1}{4}\left\{ {{{\tan }^{ - 1}}x - \frac{7}{{{{\left( {x - 1} \right)}^2}}} + \ln \left( {\frac{{\sqrt {{x^2} + 1} }}{{\left| {x - 1} \right|}}} \right)} \right\} + C\end{align}\]
 

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