8.5 | Partial Fractions


A rational function
$$
\frac{P(x)}{Q(x)}
$$
with the degree of \(P(x)\) less than the degree of \(Q(x)\), can be rewritten as a sum of fractions as follows:

  • For each factor of \(Q(x)\) of the form \((ax+b)^m\), introduce terms
    $$
    \frac{A_1}{ax +b} + \frac{A_2}{(ax+b)^2}+ \cdots +\frac{A_m}{(ax+b)^m}
    $$
  • For each factor of \(Q(x)\) of the form \((ax^2+bx+c)^m\), introduce terms
    $$
    \frac{A_1 x + B_1}{ax^2+bx+c}+\frac{A_2 x + B_2}{(ax^2+bx+c)^2}+ \cdots +\frac{A_m x + B_m}{(ax^2+bx+c)^m}
    $$

where \(A_i\)’s and \(B_i\)’s are constants. At times, for ease of notation, we will use \(A,B,C,D,…\) in place of \(A_1, B_1,\) etc… If the degree of \(P(x)\) is not less than the degree of \(Q(x)\), long divide first.

Find a partial fractions decomposition for
$$
\frac{-x^4-2 x^3-8 x^2-26 x+9}{x^5+10 x^3+9 x}
$$

$$
\frac{-x^4-2 x^3-8 x^2-26 x+9}{x^5+10 x^3+9 x} = \frac{A}{x}+\frac{B x+\text{CC}}{x^2+1}+\frac{\text{DD} x+\text{EE}}{x^2+9}
$$
for some choice of \(A,B,CC,DD,\) and \(EE\).

We clear the denominator to obtain
$$
-x^4-2 x^3-8 x^2-26 x+9 = x^4 (A+B+\text{DD})+x^2 (10 A+9 B+\text{DD})+9 A+x^3 (\text{CC}+\text{EE})+x (9
\text{CC}+\text{EE})
$$

For these two polynomials to be equal, their coefficients must be the same. That is,
$$
\begin{array}{cccccc}
A & B & 0 & \text{DD} & 0 & -1 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
10 A & 9 B & 0 & \text{DD} & 0 & -8 \\
0 & 0 & 9 \text{CC} & 0 & \text{EE} & -26 \\
9 A & 0 & 0 & 0 & 0 & 9 \\
\end{array}
$$
where the first row corresponds to the equation \(A + B + DD = -1\), the second row corresponds to the equation \(CC + EE = -2\), and so on.

Now we reduce the system:
$$
\begin{array}{cccccc}
A & B & 0 & \text{DD} & 0 & -1 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
10 A & 9 B & 0 & \text{DD} & 0 & -8 \\
0 & 0 & 9 \text{CC} & 0 & \text{EE} & -26 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
10 A & 9 B & 0 & \text{DD} & 0 & -8 \\
0 & 0 & 9 \text{CC} & 0 & \text{EE} & -26 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
0 & 9 B & 0 & \text{DD} & 0 & -18 \\
0 & 0 & 9 \text{CC} & 0 & \text{EE} & -26 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
0 & 0 & 0 & -8 \text{DD} & 0 & 0 \\
0 & 0 & 9 \text{CC} & 0 & \text{EE} & -26 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
0 & 0 & 0 & -8 \text{DD} & 0 & 0 \\
0 & 0 & 0 & 0 & -8 \text{EE} & -8 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & \text{EE} & -2 \\
0 & 0 & 0 & -8 \text{DD} & 0 & 0 \\
0 & 0 & 0 & 0 & \text{EE} & 1 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & 0 & -3 \\
0 & 0 & 0 & -8 \text{DD} & 0 & 0 \\
0 & 0 & 0 & 0 & \text{EE} & 1 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & \text{DD} & 0 & -2 \\
0 & 0 & \text{CC} & 0 & 0 & -3 \\
0 & 0 & 0 & \text{DD} & 0 & 0 \\
0 & 0 & 0 & 0 & \text{EE} & 1 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
$$
\begin{array}{cccccc}
0 & B & 0 & 0 & 0 & -2 \\
0 & 0 & \text{CC} & 0 & 0 & -3 \\
0 & 0 & 0 & \text{DD} & 0 & 0 \\
0 & 0 & 0 & 0 & \text{EE} & 1 \\
A & 0 & 0 & 0 & 0 & 1 \\
\end{array}
$$
So, we see that \(A = 1, B= -2, CC = -3, DD = 0,\) and \(EE = 1\).

Find
$$
\int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \; dx
$$
Hint: \(x^2+9\) is a factor of the denominator.


Since the degree of the numerator is less than the degree the denominator, we start by factoring the denominator completely. Long division of \(x^4-5x^3+15x^2-45x+54\) by \(x^2 + 9\) yields:
$$ x^2-5x+6$$
which factors as \((x-3)(x-2)\). So,

$$
\frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} = \frac{6x^3-20x^2+53x-87}{(x-3)(x-2)(x^2+9)}
$$

The partial fractions decomposition of the integrand will be

$$
\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9},
$$

where the constants \(A,B,C,\) and \(D\) satisfy

$$
\frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} = \frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9}.
$$

By clearing the denominator we obtain:

$$ \begin{aligned}6 & x^3 -20x^2+53x-87 \\
& = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \qquad + (Cx+D)(x-2)(x-3) \end{aligned}$$

Evaluate this equation at \(x = 1,2, 3,\) and \(4\):

$$
\begin{array}{c | c }
x & \begin{aligned}6 & x^3 -20x^2+53x-87 \\
& = A(x-2)(x^2+9)+B(x-3)(x^2+9) \\
& \; \; + (Cx+D)(x-2)(x-3) \end{aligned}\\ \hline
1 & \begin{aligned} & -48 = -10 A – 20 B + 2 C + 2D \\ \Rightarrow & -48 = -30 – 20 + 2C + 2D \\ \Rightarrow & 1 = C+D \end{aligned} \\ \hline
2 & -13 = -13B \Rightarrow B = 1 \\ \hline
3 & 54 = 18A \Rightarrow A = 3 \\ \hline
4 & \begin{aligned} & 189 = 50 A + 25 B + 8C + 2D \\ \Rightarrow & 189 = 150 + 25 + 8C +2D \\ \Rightarrow & 7= 4C+ D \end{aligned} \\
\end{array}
$$

So, \(A = 3\) and \(B = 1\). To find \(C\) and \(D\), solve the system:
$$
\begin{cases}
1 & = C &+ D \\
7 & = 4C &+ D
\end{cases}
$$

This system has solutions \(C = 2\) and \(D = -1\).

So,

$$
\begin{aligned}
& \int \frac{6x^3-20x^2+53x-87}{x^4-5x^3+15x^2-45x+54} \; dx \\
&= \int \left(\frac{A}{x-3}+\frac{B}{x-2} + \frac{Cx+D}{x^2+9} \right) \; dx\\
&= \int \left(\frac{3}{x-3}+\frac{1}{x-2} + \frac{2x-1}{x^2+9} \right) \; dx\\
&= \int \frac{3}{x-3} \; dx + \int \frac{1}{x-2} \; dx + \int \frac{2x-1}{x^2+9} \; dx \\
& = 3\ln|x-3| + \ln|x-2| + \int \frac{2x}{x^2+9} \; dx \\
& \qquad- \int \frac{1}{x^2+9} \; dx \\
& = 3 \ln|x-3| + \ln|x-2| + \ln|x^2+9| \\
& \qquad – \frac{1}{3}\tan^{-1}\left(\frac{x}{3} \right) + C
\end{aligned}
$$

E 8.5 Exercises