# 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}