Partial fraction decomposition

1.9. Partial fraction decomposition#

Integrals like

(1.193)#\[\begin{equation} \int \frac{3x+11}{x^2-x-6} \d{x} \end{equation}\]

can be evaluated by breaking them apart into simpler integrals. Let’s try to break the integrand up into the factors of its denominator:

(1.194)#\[\begin{equation} \frac{3x+11}{x^2-x-6} = \frac{3x+11}{(x-3)(x+2)} = \frac{A_1}{x-3} + \frac{A_2}{x+2} \end{equation}\]

To find \(A_1\) and \(A_2\), first cross multiply to eliminate the denominator, then combine like powers of x:

(1.195)#\[\begin{align} 3x + 11 &= A_1(x+2) + A_2(x-3) \\ 3x + 11 &= (A_1 + A_2)x + (2A_1 - 3A_2) \\ \end{align}\]

If our proposed form will work, all the coefficients of x must match on both sides, so

(1.196)#\[\begin{align} A_1 + A_2 &= 3 \\ 2A_1 - 3A_2 &= 11 \end{align}\]

This is a system of linear equations that can be solved, e.g., using a matrix technique. Form the augmented matrix then perform Gauss-Jordan elimination:

(1.197)#\[\begin{align} \begin{bmatrix} 1 & 1 & 3\\ 2 & -3 & 11 \end{bmatrix} \begin{matrix}\vphantom{R_1} \\ -2 R_1 \end{matrix} &\to \begin{bmatrix} 1 & 1 & 3\\ 0 & -5 & 5 \end{bmatrix} \begin{matrix}\vphantom{R_1} \\ \div -5 \end{matrix} \\ &\to \begin{bmatrix} 1 & 1 & 3\\ 0 & 1 & -1 \end{bmatrix} \begin{matrix} -R_2 \\ \vphantom{R_2} \end{matrix} \\ &\to \begin{bmatrix} 1 & 0 & 4\\ 0 & 1 & -1 \end{bmatrix} \end{align}\]

so \(A_1 = 4\) and \(A_2 = -1\). Once you have coefficients, you can integrate:

(1.198)#\[\begin{align} \int \frac{3x+11}{x^2-x-6} \d{x} &= \int\left(\frac{4}{x-3}-\frac{1}{x+2} \right) \d{x} \\ &= 4\ln| x-3 | - \ln|x+2| + c \end{align}\]

We can formalize this technique, called partial fraction decomposition.

To demonstrate this technique, let’s try to take the integral

(1.203)#\[\begin{equation} \int \frac{x^2-29x+5}{(x-4)^2 (x^2+3)} \d{x} \end{equation}\]

First, expand the denominator and put placeholder coefficient terms in the numerator.

(1.204)#\[\begin{equation} \frac{x^2-29x+5}{\left(x-4 \right)^2 \left(x^2+3 \right)} = \frac{A_1}{x-4} + \frac{A_2}{(x-4)^2} + \frac{A_3x+ B_3}{x^2+3} \end{equation}\]

Next, cross multiply, expand, and collect like powers of x:

(1.205)#\[\begin{align} x^2-29x+5 &= A_1 \left(x-4 \right) \left(x^2+3 \right) + A_2 \left(x^2+3 \right) \\ &\quad + \left(A_3x+B_3 \right) \left(x-4 \right)^2 \\ &= A_1 \left(x^3-4x^2+3x-12 \right) + A_2 \left(x^2+3 \right) \\ &\quad +\left(A_3x+B_3 \right)\left(x^2-8x+16 \right) \\ &=\left(A_1+A_3\right)x^3+\left(-4A_1+A_2-8A_3+B_3\right)x^2 \\ &\quad + \left(3A_1+16A_3-8B_3\right)x+\left(-12A_1+3A_2+16B_3\right) \end{align}\]

Finally, form the linear system of equations for the coefficients of x:

(1.206)#\[\begin{align} A_1+A_3 &= 0 \\ -4A_1 + A_2-8A_3 + B_3 &=1 \\ 3A_1 + 16A_3 - 8B_3 &= -29 \\ -12A_1+3A_2+16B_3 &=5 \\ \end{align}\]

These linear equations can be written using a matrix and vectors

(1.207)#\[\begin{equation} \begin{bmatrix} 1 & 0 & 1 & 0\\ -4 & 1 & -8 & 1\\ 3 & 0 & 16 & -8\\ -12 & 3 & 0 & 16 \end{bmatrix} \begin{bmatrix} A_1 \\ A_2\\ A_3\\ B_3 \end{bmatrix} = \begin{bmatrix} 0\\ 1\\ -29 \\ 5 \end{bmatrix} \end{equation}\]

Solving numerically gives \(A_1 = 1\), \(A_2 = -5\), \(A_3 = -1\), and \(B_3 =2\). Finally,

(1.208)#\[\begin{align} \int &\frac{x^2-29x+5}{(x-4)^2 (x^2+3)} \d{x} \\ &= \int\Biggl(\frac{1}{x-4}-\frac{5}{(x-4)^2} -\frac{x}{x^2+3}+\frac{2}{x^2+3}\Biggr) \d{x} \\ &= \ln|x-4|+\frac{5}{x-4} -\frac{1}{2} \ln|x^2 + 3| + \frac{2}{\sqrt{3}} \arctan\left( \frac{x}{\sqrt{3}} \right) + c \end{align}\]

using u substitution in

(1.209)#\[\begin{equation} \int \frac{\d{x}}{1+x^2} = \arctan x \end{equation}\]

to evaluate the last integral.

1.9.1. Heaviside cover-up method#

The Heaviside cover-up method is a simplified approach to finding the coefficients in partial fraction decomposition for linear factors. Its main advantage is avoiding cross-multiplication, which can be time-consuming, by covering up a factor and substituting the appropriate value into the expression.

We will demonstrate the approach by example. Given the decomposition:

(1.210)#\[\begin{equation} \frac{3x+11}{(x-3)(x+2)} = \frac{A_1}{x-3} + \frac{A_2}{x+2} \end{equation}\]

Multiply by both sides by the first factor and take the limit to find \(A_1\):

(1.211)#\[\begin{align} \lim_{x\to 3} \left( \frac{3x+11}{x+2} \right) &= \lim_{x \to 3} \left( A_1 + \frac{A_2(x-3)}{x+2} \right) \\ \frac{3(3)+11}{3+2} &= A_1 \\ A_1 &= 4 \end{align}\]

You can do the same for the second factor to get \(A_2\):

(1.212)#\[\begin{align} \lim_{x\to -2} \left( \frac{3x+11}{x-3}\right) &= \lim_{x\to -2} \left( \frac{A_1(x+2)}{x-3} + A_2 \right) \\ \frac{3(-2)+11}{-2-3} &= A_2 \\ A_2 &= -1 \end{align}\]

This procedure can be implemented conveniently by “covering up” the relevant factor in the original fraction!

Note that this procedure only works for the highest powers of linear factors, but it can be used in combination with other techniques to more easily obtain the remaining factors.

For example, given:

(1.213)#\[\begin{equation} \frac{x^2-29x+5}{(x-4)^2 (x^2 +3)} = \frac{A_1}{x-4} + \frac{A_2}{(x-4)^2} + \frac{A_3x + B_3}{x^2 +3} \end{equation}\]

\(A_2\) can be solved for using the Heaviside cover-up method because it corresponds to the highest power of the factor \(x-4\).

(1.214)#\[\begin{equation} A_2 = \frac{4^2 - 29 \cdot 4 + 5}{4^2 + 3} = -5 \end{equation}\]

Cross-multiplication must be used to solve for \(A_1\), \(A_3\), and \(B_3\).

1.9.2. Polynomial division#

Polynomial division is a method used to simplify rational functions when the degree of the numerator \(P\) is higher than the degree of the denominator \(Q\). The remaineder from the division should then be suitable for partial fraction decomposition.

For example, to integrate:

(1.215)#\[\begin{equation} \int \frac{x^4 - 5x^3 + 6x^2 -18}{x^3 -3x^2} \d{x} \end{equation}\]

Partial fraction decomposition seems useful, but the degree of the numerator P is 4 but the degree of denominator Q is 3, so we must carry out polynomial division first.

\[\begin{split} \begin{array}{r} x-2 \phantom{5x^3+6x^2-18}\\ x^3-3x^2{\overline{\smash{\big)}\, x^4-5x^3+6x^2-18}} \\ \underline{-~(x^4-3x^3)}\phantom{+6x^2-18} \\ -2x^3+6x^2-18 \\ \underline{-~(-2x^3+6x^2)}\phantom{-18} \\ -18 \\ \end{array} \end{split}\]

The result is

(1.216)#\[\begin{equation} \frac{x^4 - 5x^3 + 6x^2 -18}{x^3 -3x^2} = x - 2 - \frac{18}{x^3-3x^2} \end{equation}\]

Use partial fraction decomposition for the fractional term:

(1.217)#\[\begin{equation} \frac{-18}{x^2 (x-3)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x-3} \end{equation}\]

Find \(A_2\) and \(A_3\) using the Heaviside cover-up method:

(1.218)#\[\begin{align} A_2 &= \frac{-18}{0-3} = 6 \\ A_3 &= \frac{-18}{3^2} = -2 \end{align}\]

Find \(A_1\) by substituting \(A_2\) and \(A_3\), then cross-multiplying:

(1.219)#\[\begin{equation} -18 = A_1 x (x-3) + 6(x-3) - 2x^2 \end{equation}\]

We could collect powers of x, but plugging in \(x=1\) then solving for \(A_1\) is a little easier:

(1.220)#\[\begin{align} -18 &= A_1 \cdot 1 \cdot -2 + 6 \cdot -2 - 2 = -2A_1 - 14 \\ A_1 &= 2 \end{align}\]

Last, integrate:

(1.221)#\[\begin{align} &\int \frac{x^4 - 5x^3 + 6x^2 -18}{x^3 -3x^2} \d{x} \\ &= \int \left(x - 2 + \frac{2}{x} + \frac{6}{x^2} - \frac{2}{x-3} \right) \d{x} \\ &= \frac{x^2}{2} -2x + 2\ln|x| - \frac{6}{x} - 2\ln|x-3| + c \end{align}\]

1.9.3. Skill builder problems#

Integrate the following:

  1. \(\displaystyle \int \frac{2x+16}{x^2-16} \d{x}\)

    Solution

    Fractional decomposition:

    (1.222)#\[\begin{equation} \frac{2x+16}{(x+4)(x-4)} = \frac{A_1}{x+4} + \frac{A_2}{x-4} \end{equation}\]

    Cover up:

    (1.223)#\[\begin{align} A_1 &= \frac{2(-4)+16}{-4-4} = -1 \\ A_2 &= \frac{2(4)+16}{4+4} = 3 \end{align}\]

    so

    (1.224)#\[\begin{align} \int\frac{2x+16}{x^2-16} \d{x} &= \int\left( -\frac{1}{x+4}+\frac{3}{x-4} \right) \d{x} \\ &= -\ln|x+4| + 3\ln|x-4| + c \end{align}\]

  2. \(\displaystyle \int \frac{8}{x^2+2x} \d{x}\)

    Solution

    Fractional decomposition:

    (1.225)#\[\begin{equation} \frac{8}{x(x+2)} = \frac{A_1}{x} + \frac{A_2}{x+2} \end{equation}\]

    Cover up:

    (1.226)#\[\begin{align} A_1 &= \frac{8}{0+2} = 4 \\ A_2 &= \frac{8}{-2} = -4 \end{align}\]

    so

    (1.227)#\[\begin{align} \int\frac{8}{x^2+2x} \d{x} &= \int\left(\frac{4}{x} - \frac{4}{x+2}\right) \d{x} \\ &= 4\ln|x| - 4\ln|x+2| + c \end{align}\]

  3. \(\displaystyle \int \frac{5+12x^2-x^3}{x^2(x-9)(x-1)} \d{x}\)

    Solution

    Fractional decomposition

    (1.228)#\[\begin{equation} \frac{5+12x^2-x^3}{x^2(x-9)(x-1)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x-9} + \frac{A_4}{x-1} \end{equation}\]

    Cover up:

    (1.229)#\[\begin{align} A_2 &= \frac{5+12(0^2)-0^3}{(0-9)(0-1)} = \frac{5}{9} \\ A_3 &= \frac{5+12(9^2)-9^3}{9^2(9-1)} = \frac{31}{81} \\ A_4 &= \frac{5+12(1^2)-1^3}{1^2(1-9)} = -2 \end{align}\]

    To obtain \(A_1\), plugin \(x = -1\)

    (1.230)#\[\begin{align} \frac{9}{10} &= -A_1 + \frac{5}{9} - \frac{1}{10}\cdot\frac{31}{81} + 1 \\ A_1 &= \frac{50}{81} \end{align}\]

    so

    (1.231)#\[\begin{align} \int\frac{5+12x^2-x^3}{(x^2)(x-9)(x-1)} \d{x} & =\int\Biggl(\frac{50}{81}\frac{1}{x} + \frac{5}{9}\frac{1}{x^2} + \frac{31}{81}\frac{1}{x-9} - \frac{2}{x-1}\Biggr) \d{x} \\ &= \frac{50}{81}\ln|x| - \frac{5}{9}\frac{1}{x} + \frac{31}{81}\ln|x-5| - 2\ln|x-1| + c \end{align}\]
Contents