Calculus

Integration of Functions

Integration of Functions Logo

Partial Fraction Decomposition

  • The Idea behind Partial Fraction Decomposition

    A rational function is a function of the form

    \[{f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}},}\]

    where \(P\left( x \right)\) and \(Q\left( x \right)\) are polynomials in \(x\) with \(Q\left( x \right) \not\equiv 0.\)

    Rational fractions can be integrated by splitting into partial fractions.

    Consider the addition of two rational fractions:

    \[{\frac{2}{{x + 1}} + \frac{3}{{x + 4}} }={ \frac{{2\left( {x + 4} \right) + 3\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 4} \right)}} }={ \frac{{\color{red}{2x} + \color{blue}8 + \color{red}{3x} + \color{blue}3}}{{\color{darkgreen}{x^2} + \color{red}x + \color{red}{4x} + \color{blue}4}} }={ \frac{{\color{red}{5x} + \color{blue}{11}}}{{\color{darkgreen}{x^2} + \color{red}{5x} + \color{blue}4}}.}\]

    Partial Fraction Decomposition (PFD) is the reverse of the procedure. It allows to decompose a single rational function into a sum of simpler rational functions.

    Considering the previous example, the partial fraction decomposition has the form

    \[{\frac{{5x + 11}}{{{x^2} + 5x + 4}} }={ \frac{2}{{x + 1}} + \frac{3}{{x + 4}}.}\]

    Partial fraction decomposition is used to integrate rational functions.

    The given method involves the following basic steps:

    1. Check to make sure the fraction is a proper rational function.
    2. Factor the polynomial \({Q\left( x \right)}\) in the denominator into linear and/or quadratic factors.
    3. Decompose the fraction into a sum of partial fractions.

    Let’s look at each step in detail.

    Step \(1.\) Reducing an Improper Fraction

    In a proper rational expression \(\large{\frac{{P\left( x \right)}}{{Q\left( x \right)}}}\normalsize,\) the degree of the numerator \({P\left( x \right)}\) is less than the degree of the denominator \({Q\left( x \right)}.\)

    If \(\large{\frac{{P\left( x \right)}}{{Q\left( x \right)}}}\normalsize\) is improper, we divide the numerator \({P\left( x \right)}\) by the denominator \({Q\left( x \right)}\) to obtain

    \[{\frac{{P\left( x \right)}}{{Q\left( x \right)}} = F\left( x \right) + \frac{{R\left( x \right)}}{{Q\left( x \right)}},}\]

    where \(F\left( x \right)\) is a polynomial and \(\large{\frac{{R\left( x \right)}}{{Q\left( x \right)}}}\normalsize\) is a proper fraction.

    Step \(2.\) Factoring the Denominator \(Q\left( x \right)\) into Linear/Quadratic Factors

    Any polynomial with real coefficients can be factored into a product of linear and/or irreducible quadratic factors.

    Step \(3.\) Splitting the Rational Function into a Sum of Partial Fractions

    The partial fraction decomposition of the rational function \(\large{\frac{{R\left( x \right)}}{{Q\left( x \right)}}}\normalsize\) depends on the factors of the denominator \({Q\left( x \right)}.\)

    \(1.\) The denominator is a product of distinct linear factors.

    For each distinct factor \(ax + b,\) the sum of partial fractions includes a term of the form \(\large{\frac{A}{{ax + b}}}\normalsize.\)

    \(2.\) The denominator is a product of linear factors, some of which are repeated.

    For each repeated linear factor \({\left( {ax + b} \right)^\alpha },\) the sum of partial fractions includes \(\alpha\) terms of the form

    \[{\frac{{{A_1}}}{{ax + b}}} + {\frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} }+ \ldots + {\frac{{{A_\alpha }}}{{{{\left( {ax + b} \right)}^\alpha }}}.}\]

    \(3.\) The denominator has distinct irreducible quadratic factors.

    For each distinct factor \(a{x^2} + bx + c,\) the sum of partial fractions includes a term \(\large{\frac{{Ax + B}}{{a{x^2} + bx + c}}}\normalsize.\)

    \(4.\) The denominator has a repeated irreducible quadratic factor.

    For each repeated quadratic factor \({\left( {a{x^2} + bx + c} \right)^\mu },\) the sum of partial fractions includes \(\mu\) terms of the form

    \[{\frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} }+{ {\frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}}} + \ldots }+{ \frac{{{A_\mu }x + {B_\mu }}}{{{{\left( {a{x^2} + bx + c} \right)}^\mu }}}.}\]

    Once the partial fraction decomposition is determined, the unknown coefficients can be found by clearing fractions and equating the coefficients of similar powers on the two sides. The resulting system must always have a unique solution.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Decompose the fraction \(\large{\frac{6}{{\left( {x + 2} \right)\left( {x – 4} \right)}}}\normalsize.\)

    Example 2

    Decompose the fraction \(\large{\frac{3x}{{\left( {x + 1} \right)\left( {x – 2} \right)}}}\normalsize.\)

    Example 3

    Find a partial fraction decomposition for \(\large{\frac{{x + 4}}{{{x^2} – 7x + 10}}}\normalsize.\)

    Example 4

    Find a partial fraction decomposition for \(\large{\frac{{2x – 3}}{{{x^2} + 7x + 6}}}\normalsize.\)

    Example 5

    Find a partial fraction decomposition for \(\large{\frac{{x – 2}}{{{x^2}\left( {x + 1} \right)}}}\normalsize.\)

    Example 6

    Find a partial fraction decomposition for \(\large{\frac{1}{{x{{\left( {2x + 1} \right)}^2}}}}\normalsize.\)

    Example 7

    Determine a partial fraction decomposition for the function \(\large{\frac{{15x}}{{\left( {x + 1} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}}\normalsize.\)

    Example 8

    Determine a partial fraction decomposition for the function \(\large{\frac{{{x^2} + 2}}{{{x^3} – 3{x^2} + 2x}}}\normalsize.\)

    Example 9

    Decompose \(\large{\frac{6}{{x\left( {{x^2} + x + 3} \right)}}}\normalsize\) using the partial fractions.

    Example 10

    Expand the rational function \({\large{\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x – 1} \right)}^2}}}}}\normalsize\) into partial fractions.

    Example 11

    Decompose \(\large{\frac{{{x^2} – 6x}}{{\left( {x – 1} \right)\left( {{x^2} + 2x + 2} \right)}}}\normalsize\) using the partial fractions.

    Example 1.

    Decompose the fraction \(\large{\frac{6}{{\left( {x + 2} \right)\left( {x – 4} \right)}}}\normalsize.\)

    Solution.

    The factors of the denominator are linear. Therefore, the numerators of the partial fractions will be constants. The partial fraction decomposition has the form

    \[{\frac{6}{{\left( {x + 2} \right)\left( {x – 4} \right)}} }={ \frac{A}{{x + 2}} + \frac{B}{{x – 4}}.}\]

    We add the partial fractions on the right side:

    \[{\frac{A}{{x + 2}} + \frac{B}{{x – 4}} }={ \frac{{A\left( {x – 4} \right) + B\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x – 4} \right)}} }={ \frac{{Ax – 4A + Bx + 2B}}{{\left( {x + 2} \right)\left( {x – 4} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {2B – 4A} \right)}}{{\left( {x + 2} \right)\left( {x – 4} \right)}}.}\]

    This expression is equal to the original fraction, that is

    \[{\frac{6}{{\left( {x + 2} \right)\left( {x – 4} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {2B – 4A} \right)}}{{\left( {x + 2} \right)\left( {x – 4} \right)}}.}\]

    Since these fractions have the same denominators, we can equate their numerators:

    \[6 = \left( {A + B} \right)x + \left( {2B – 4A} \right).\]

    Hence

    \[{\left\{ \begin{array}{l} A + B = 0\\ 2B – 4A = 6 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = – B\\ 2B + 4B = 6 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = – B\\ 6B = 6 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = – 1\\ B = 1 \end{array} \right..}\]

    The final answer is given by

    \[{\frac{6}{{\left( {x + 2} \right)\left( {x – 4} \right)}} }={ \frac{1}{{x – 4}} – \frac{1}{{x + 2}}.}\]

    Example 2.

    Decompose the fraction \(\large{\frac{3x}{{\left( {x + 1} \right)\left( {x – 2} \right)}}}\normalsize.\)

    Solution.

    Since the factors of the denominator are linear, the partial fraction decomposition is written in the form

    \[{\frac{{3x}}{{\left( {x + 1} \right)\left( {x – 2} \right)}} }={ \frac{A}{{x + 1}} + \frac{B}{{x – 2}}.}\]

    Simplifying the right side, we have

    \[{\frac{A}{{x + 1}} + \frac{B}{{x – 2}} }={ \frac{{A\left( {x – 2} \right) + B\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x – 2} \right)}} }={ \frac{{Ax – 2A + Bx + B}}{{\left( {x + 1} \right)\left( {x – 2} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {B – 2A} \right)}}{{\left( {x + 1} \right)\left( {x – 2} \right)}}.}\]

    This expression is equal to the original fraction:

    \[{\frac{{3x}}{{\left( {x + 1} \right)\left( {x – 2} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {B – 2A} \right)}}{{\left( {x + 1} \right)\left( {x – 2} \right)}}.}\]

    The fractions on both sides have the same denominators, so we equate their numerators:

    \[3x = \left( {A + B} \right)x + \left( {B – 2A} \right).\]

    Then

    \[{\left\{ \begin{array}{l} A + B = 3\\ B – 2A = 0 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A + 2A = 3\\ B = 2A \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} 3A = 3\\ B = 2A \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 1\\ B = 2 \end{array} \right..}\]

    Thus, the partial fraction decomposition has the form

    \[{\frac{{3x}}{{\left( {x + 1} \right)\left( {x – 2} \right)}} }={ \frac{1}{{x + 1}} + \frac{2}{{x – 2}}.}\]

    Example 3.

    Find a partial fraction decomposition for \(\large{\frac{{x + 4}}{{{x^2} – 7x + 10}}}\normalsize.\)

    Solution.

    Calculate the discriminant of the quadratic function in the denominator:

    \[{D = {\left( { – 7} \right)^2} – 4 \cdot 10 }={ 9 \gt 0.}\]

    As the discriminant is positive, we can determine the roots of the quadratic function and factor it:

    \[{{x_{1,2}} = \frac{{7 \pm \sqrt 9 }}{2} }={ \frac{{7 \pm 3}}{2} }={ 2,\,5.}\]

    Hence

    \[{{x^2} – 7x + 10 }={ \left( {x – 2} \right)\left( {x – 5} \right),}\]

    and the original fraction is written as

    \[{\frac{{x + 4}}{{{x^2} – 7x + 10}} }={ \frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 5} \right)}}.}\]

    The denominator contains linear factors, so the partial fraction decomposition has the form

    \[{\frac{{x + 4}}{{{x^2} – 7x + 10}} }={ \frac{A}{{x – 2}} + \frac{B}{{x – 5}}.}\]

    Determine the unknown coefficients \(A\) and \(B:\)

    \[{\frac{A}{{x – 2}} + \frac{B}{{x – 5}} }={ \frac{{A\left( {x – 5} \right) + B\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x – 5} \right)}} }={ \frac{{Ax – 5A + Bx – 2B}}{{\left( {x – 2} \right)\left( {x – 5} \right)}} }={ \frac{{\left( {A + B} \right)x – 5A – 2B}}{{\left( {x – 2} \right)\left( {x – 5} \right)}}.}\]

    Then we can write

    \[{\frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 5} \right)}} }={ \frac{{\left( {A + B} \right)x – 5A – 2B}}{{\left( {x – 2} \right)\left( {x – 5} \right)}}.}\]

    Equate the numerators:

    \[x + 4 = \left( {A + B} \right)x – 5A – 2B.\]

    Hence

    \[{\left\{ \begin{array}{l} A + B = 1\\ – 5A – 2B = 4 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 1 – B\\ – 5\left( {1 – B} \right) – 2B = 4 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 1 – B\\ – 5 + 5B – 2B = 4 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 1 – B\\ 3B = 9 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = – 2\\ B = 3 \end{array} \right..}\]

    The final answer is given by

    \[{\frac{{x + 4}}{{{x^2} – 7x + 10}} }={ \frac{3}{{x – 5}} – \frac{2}{{x – 2}}.}\]

    Example 4.

    Find a partial fraction decomposition for \(\large{\frac{{2x – 3}}{{{x^2} + 7x + 6}}}\normalsize.\)

    Solution.

    The discriminant of the quadratic function in the denominator is positive:

    \[{D = {7^2} – 4 \cdot 6 }={ 25 \gt 0,}\]

    so we can factor it:

    \[{{x_{1,2}} = \frac{{ – 7 \pm \sqrt {25} }}{2} }={ \frac{{ – 7 \pm 5}}{2} }={ – 1,\, – 6;}\]

    \[{ \Rightarrow {x^2} + 7x + 6 }={ \left( {x + 1} \right)\left( {x + 6} \right).}\]

    Then the fraction can be written as

    \[{\frac{{2x – 3}}{{{x^2} + 7x + 6}} }={ \frac{{2x – 3}}{{\left( {x + 1} \right)\left( {x + 6} \right)}}.}\]

    The denominator contains linear factors. Therefore, the partial fraction decomposition is given by

    \[{\frac{{2x – 3}}{{\left( {x + 1} \right)\left( {x + 6} \right)}} }={ \frac{A}{{x + 1}} + \frac{B}{{x + 6}}.}\]

    Simplify the right-hand side:

    \[{\frac{A}{{x + 1}} + \frac{B}{{x + 6}} }={ \frac{{A\left( {x + 6} \right) + B\left( {x + 1} \right)}}{{\left( {x + 6} \right)\left( {x + 1} \right)}} }={ \frac{{Ax + 6A + Bx + B}}{{\left( {x + 6} \right)\left( {x + 1} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {6A + B} \right)}}{{\left( {x + 6} \right)\left( {x + 1} \right)}}.}\]

    As

    \[{\frac{{2x – 3}}{{\left( {x + 1} \right)\left( {x + 6} \right)}} }={ \frac{{\left( {A + B} \right)x + \left( {6A + B} \right)}}{{\left( {x + 6} \right)\left( {x + 1} \right)}},}\]

    we can write

    \[{2x – 3 }={ \left( {A + B} \right)x + \left( {6A + B} \right).}\]

    This yields:

    \[{\left\{ \begin{array}{l} A + B = 2\\ 6A + B = – 3 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 2 – B\\ 6\left( {2 – B} \right) + B = – 3 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 2 – B\\ 12 – 6B + B = – 3 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = 2 – B\\ 5B = 15 \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} A = – 1\\ B = 3 \end{array} \right..}\]

    So the decomposition of the fraction has the form

    \[{\frac{{2x – 3}}{{{x^2} + 7x + 6}} }={ \frac{3}{{x + 6}} – \frac{1}{{x + 1}}.}\]

    Example 5.

    Find a partial fraction decomposition for \(\large{\frac{{x – 2}}{{{x^2}\left( {x + 1} \right)}}}\normalsize.\)

    Solution.

    The first factor \(x\) in the denominator is repeated twice, so it produces two partial fractions. The resulting decomposition has the form

    \[{\frac{{x – 2}}{{{x^2}\left( {x + 1} \right)}} }={ \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}}.}\]

    First we simplify the right-hand side of this expression:

    \[{\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} }={ \frac{{Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2}}}{{{x^2}\left( {x + 1} \right)}} }={ \frac{{A{x^2} + Ax + Bx + B + C{x^2}}}{{{x^2}\left( {x + 1} \right)}} }={ \frac{{\left( {A + C} \right){x^2} + \left( {A + B} \right)x + B}}{{{x^2}\left( {x + 1} \right)}}.}\]

    The last fraction is equal to the original one:

    \[{\frac{{x – 2}}{{{x^2}\left( {x + 1} \right)}} }={ \frac{{\left( {A + C} \right){x^2} + \left( {A + B} \right)x + B}}{{{x^2}\left( {x + 1} \right)}}.}\]

    Then we equate the numerators of both sides:

    \[{x – 2 }={ \left( {A + C} \right){x^2} + \left( {A + B} \right)x + B.}\]

    Hence we obtain the following system of equations:

    \[\left\{ \begin{array}{l} A + C = 0\\ A + B = 1\\ B = – 2 \end{array} \right..\]

    Solving it, we find the unknown coefficients:

    \[\left\{ \begin{array}{l} C = – 3\\ A = 3\\ B = – 2 \end{array} \right..\]

    So, the partial fraction decomposition is written as

    \[{\frac{{x – 2}}{{{x^2}\left( {x + 1} \right)}} }={ \frac{3}{x} – \frac{2}{{{x^2}}} – \frac{3}{{x + 1}}.}\]

    Example 6.

    Find a partial fraction decomposition for \(\large{\frac{1}{{x{{\left( {2x + 1} \right)}^2}}}}\normalsize.\)

    Solution.

    The second factor \(2x + 1\) in the denominator is repeated twice. Hence it produces two partial fractions, so the decomposition has the form

    \[{\frac{1}{{x{{\left( {2x + 1} \right)}^2}}} }={ \frac{A}{{2x + 1}} + \frac{B}{{{{\left( {2x + 1} \right)}^2}}} + \frac{C}{x}.}\]

    We combine the partial fractions on the right side into a single fraction and then equate the numerators on both sides. This gives the following equation:

    \[{1 = \left( {2A + 4C} \right){x^2} }+{ \left( {A + B + 4C} \right)x }+{ C,}\]

    so we can write

    \[\left\{ \begin{array}{l} 2A + 4C = 0\\ A + B + 4C = 0\\ C = 1 \end{array} \right..\]

    The system has the solution

    \[\left\{ \begin{array}{l} A = – 2\\ B = – 2\\ C = 1 \end{array} \right..\]

    Hence, the partial fraction decomposition of the initial rational function is given by

    \[{\frac{1}{{x{{\left( {2x + 1} \right)}^2}}} }={ \frac{1}{x} – \frac{2}{{2x + 1}} – \frac{2}{{{{\left( {2x + 1} \right)}^2}}}.}\]

    Example 7.

    Determine a partial fraction decomposition for the function \(\large{\frac{{15x}}{{\left( {x + 1} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}}\normalsize.\)

    Solution.

    The partial fraction decomposition has the form

    \[{\frac{{15x}}{{\left( {x + 1} \right)\left( {x – 2} \right)\left( {x + 3} \right)}} }={ \frac{A}{{x + 1}} + \frac{B}{{x – 2}} + \frac{C}{{x + 3}}.}\]

    We combine the partial fractions on the right side into a single fraction and then equate the numerators on both sides. This yields:

    \[{15x = A\left( {x – 2} \right)\left( {x + 3} \right) }+{ B\left( {x + 1} \right)\left( {x + 3} \right) }+{ C\left( {x + 1} \right)\left( {x – 2} \right),}\]

    \[{15x = A\left( {{x^2} + x – 6} \right) }+{ B\left( {{x^2} + 4x + 3} \right) }+{ C\left( {{x^2} – x – 2} \right),}\]

    \[{15x = \left( {A + B + C} \right){x^2} }+{ \left( {A + 4B – C} \right)x }+{ \left( { – 6A + 3B – 2C} \right).}\]

    Hence we obtain the following system of equations:

    \[\left\{ \begin{array}{l} A + B + C = 0\\ A + 4B – C = 15\\ – 6A + 3B – 2C = 0 \end{array} \right..\]

    Solving it, we find the unknown coefficients:

    \[A = 5,\;B = 4,\;C = – 9.\]

    The partial fraction decomposition is given by

    \[{\frac{{15x}}{{\left( {x + 1} \right)\left( {x – 2} \right)\left( {x + 3} \right)}} }={ \frac{5}{{x + 1}} + \frac{4}{{x – 2}} – \frac{9}{{x + 3}}.}\]

    Example 8.

    Determine a partial fraction decomposition for the function \(\large{\frac{{{x^2} + 2}}{{{x^3} – 3{x^2} + 2x}}}\normalsize.\)

    Solution.

    First we factor the cubic function in the denominator:

    \[{{x^3} – 3{x^2} + 2x }={ x\left( {{x^2} – 3x + 2} \right) }={ x\left( {x – 1} \right)\left( {x – 2} \right).}\]

    As the denominator contains only linear factors, the decomposition has the form

    \[{\frac{{{x^2} + 2}}{{{x^3} – 3{x^2} + 2x}} }={ \frac{A}{x} + \frac{B}{{x – 1}} + \frac{C}{{x – 2}}.}\]

    Combining the partial fractions on the right side into a single fraction and equating the numerators on both sides, we have:

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

    \[{{x^2} + 2 }={ A\left( {{x^2} – 3x + 2} \right) }+{ B\left( {{x^2} – 2x} \right) }+{ C\left( {{x^2} – x} \right),}\]

    \[{{x^2} + 2 }={ \left( {A + B + C} \right){x^2} }+{ \left( { – 3A – 2B – C} \right)x }+{ 2A.}\]

    The system of equations for the unknown coefficients is written as

    \[\left\{ \begin{array}{l} A + B + C = 1\\ – 3A – 2B – C = 0\\ 2A = 2 \end{array} \right..\]

    It has the following solution:

    \[\left\{ \begin{array}{l} A = 1\\ B = – 3\\ C = 3 \end{array} \right.,\]

    so the decomposition of the original rational function is expressed by the equation:

    \[{\frac{{{x^2} + 2}}{{{x^3} – 3{x^2} + 2x}} }={ \frac{1}{x} – \frac{3}{{x – 1}} + \frac{3}{{x – 2}}.}\]

    Example 9.

    Decompose \(\large{\frac{6}{{x\left( {{x^2} + x + 3} \right)}}}\normalsize\) using the partial fractions.

    Solution.

    Given the irreducible quadratic factor \({{x^2} + x + 3}\) in the denominator, we write the partial fraction expansion in the form

    \[{\frac{6}{{x\left( {{x^2} + x + 3} \right)}} }={ \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + x + 3}}.}\]

    The equation for the unknown coefficients is given by

    \[{6 = A\left( {{x^2} + x + 3} \right) }+{ B{x^2} }+{ Cx,}\]

    or

    \[{6 = \left( {A + B} \right){x^2} }+{ \left( {A + C} \right)x }+{ 3A.}\]

    We obtain the system of equations:

    \[\left\{ \begin{array}{l} A + B = 0\\ A + C = 0\\ 3A = 6 \end{array} \right..\]

    This yields:

    \[A = 2,\;B = – 2,\;C = – 2.\]

    Hence, the partial fraction decomposition is written as

    \[{\frac{6}{{x\left( {{x^2} + x + 3} \right)}} }={ \frac{2}{x} – \frac{{2x + 2}}{{{x^2} + x + 3}}.}\]

    Example 10.

    Expand the rational function \({\large{\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x – 1} \right)}^2}}}}}\normalsize\) into partial fractions.

    Solution.

    Note that the denominator \(Q\left( x \right)\) contains the irreducible quadratic factor \({{x^2} + x + 2}\) and the linear factor \({{{\left( {x – 1} \right)}^2}}\) with multiplicity \(2.\) Hence, the partial fraction decomposition is written as

    \[{\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x – 1} \right)}^2}}} }={ \frac{{Ax + B}}{{{x^2} + x + 2}} }+{ \frac{C}{{x – 1}} }+{ \frac{D}{{{{\left( {x – 1} \right)}^2}}}.}\]

    To determine the unknown coefficients \(A, B, C, D\) we clear fractions multiplying both sides by the denominator. This gives

    \[{16 = \left( {Ax + B} \right){\left( {x – 1} \right)^2} }+{ C\left( {x – 1} \right)\left( {{x^2} + x + 2} \right) }+{ D\left( {{x^2} + x + 2} \right),}\]

    \[{16 = \left( {Ax + B} \right)\left( {{x^2} – 2x + 1} \right) }+{ \left( {Cx – C} \right)\left( {{x^2} + x + 2} \right) }+{ D{x^2} + Dx + 2D,}\]

    \[\require{cancel}{\color{blue}{16} = \color{magenta}{A{x^3}} }+{ \color{darkgreen}{B{x^2}} }-{ \color{darkgreen}{2A{x^2}} }-{ \color{red}{2Bx} }+{ \color{red}{Ax} }+{ \color{blue}B }+{ \color{magenta}{C{x^3}} }-{ \cancel{\color{darkgreen}{C{x^2}}} }+{ \cancel{\color{darkgreen}{C{x^2}}} }-{ \color{red}{Cx} }+{ \color{red}{2Cx} }-{ \color{blue}{2C} }+{ \color{darkgreen}{D{x^2}} }+{ \color{red}{Dx} }+{ \color{blue}{2D},}\]

    \[{\color{blue}{16} }={ \left( {\color{magenta}A + \color{magenta}C} \right)\color{magenta}{x^3} }+{ \left( { – \color{darkgreen}{2A} + \color{darkgreen}B + \color{darkgreen}D} \right)\color{darkgreen}{x^2} }+{ \left( {\color{red}A – \color{red}{2B} + \color{red}C + \color{red}D} \right)\color{red}x }+{ \color{blue}B – \color{blue}{2C} + \color{blue}{2D}.}\]

    By equating coefficients of similar powers on both sides, we have

    \[{\left\{ \begin{align} \color{magenta}A + \color{magenta}C & = \color{magenta}0\\ – \color{darkgreen}{2A} + \color{darkgreen}B + \color{darkgreen}D & = \color{darkgreen}0\\ \color{red}A – \color{red}{2B} + \color{red}{C} + \color{red}D & = \color{red}0\\ \color{blue}B – \color{blue}{2C} + \color{blue}{2D} & = \color{blue}{16} \end{align} \right..}\]

    We solve the resulting system and find

    \[{A = 3,\;\;}\kern0pt{B = 2,\;\;}\kern0pt{C = – 3,\;\;}\kern0pt{D = 4.}\]

    Hence, the partial fraction decomposition of the rational function is given by

    \[{\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x – 1} \right)}^2}}} }={ \frac{{3x + 2}}{{{x^2} + x + 2}} }-{ \frac{3}{{x – 1}} }+{ \frac{4}{{{{\left( {x – 1} \right)}^2}}}.}\]

    Example 11.

    Decompose \(\large{\frac{{{x^2} – 6x}}{{\left( {x – 1} \right)\left( {{x^2} + 2x + 2} \right)}}}\normalsize\) using the partial fractions.

    Solution.

    The denominator has the irreducible quadratic factor \({{x^2} + 2x + 2}.\) Hence, the partial fraction decomposition has the form

    \[{\frac{{{x^2} – 6x}}{{\left( {x – 1} \right)\left( {{x^2} + 2x + 2} \right)}} }={ \frac{A}{{x – 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 2}}.}\]

    By equating the numerators on both sides of the equation, we have

    \[{{x^2} – 6x }={ A\left( {{x^2} + 2x + 2} \right) }+{ \left( {Bx + C} \right)\left( {x – 1} \right),}\]

    \[{{x^2} – 6x }={ A{x^2} }+{ 2Ax }+{ 2A }+{ B{x^2} }+{ Cx }-{ Bx }-{ C,}\]

    \[{{x^2} – 6x }={ \left( {A + B} \right){x^2} }+{ \left( {2A + C – B} \right)x }+{ \left( {2A – C} \right).}\]

    We get the following system of equations:

    \[\left\{ \begin{array}{l} A + B = 1\\ 2A + C – B = – 6\\ 2A – C = 0 \end{array} \right..\]

    It has the solution

    \[A = -1,\;B = 2,\;C = – 2.\]

    Hence, the partial fraction decomposition is given by

    \[{\frac{{{x^2} – 6x}}{{\left( {x – 1} \right)\left( {{x^2} + 2x + 2} \right)}} }={ \frac{{2x – 2}}{{{x^2} + 2x + 2}} – \frac{1}{{x – 1}}.}\]