Primitives of Rational Functions (Partial Fraction Decomposition)

This part deals with only one subject: How to find primitives of rational functions, i.e. of functions of the type

f = \frac{r}{s}

with polynomials r and s. Without restriction we assume s to be non constant and normalized.

On intervals, for example between two adjacent zeros of s, f has a primitive as it is continuous. There is a concept to calculate these primitives which is essentially related to two basic theorems on polynomials:

Fundamental Theorem of Algebra

Each non constant normalized polynomial s on $ℝ$ completely decomposes into a product of linear and indecomposable quadratic polynomials:

$s = {(X - a_{1})}^{l_{1}} \cdot \dots \cdot {(X - a_{j})}^{l_{j}} \cdot (X^{2} + p_{1} X + q_{1})^{n_{1}} \cdot \dots \cdot (X^{2} + p_{k} X + q_{k})^{n_{k}}$ [1]

where the discriminant of a quadratic polynomial controls its decomposability:

$X^{2} + p_{i} X + q_{i}$ is indecomposable $\Leftrightarrow D_{i} = \frac{p_{i}^{2}}{4} - q_{i} < 0$

So we know that $X^{2} + p_{i} X + q_{i} (0) = q_{i} > 0$ and that, consequently, all values of $X^{2} + p_{i} X + q_{i}$ are positive. (Continuous functions with no zeros do not vary their algebraic sign on intervals!)
Partial Fraction Theorem

Based on the decomposition [1] each rational function with a non constant mormalized denominator s is representable as a sum:

$\begin{array}{l} \frac{r}{s} = t & + \frac{c_{11}}{X - a_{1}} + \frac{c_{12}}{{(X - a_{1})}^{2}} + \dots + \frac{c_{1 l_{1}}}{{(X - a_{1})}^{l_{1}}} \\ + \dots \\ + \frac{c_{j 1}}{X - a_{j}} + \frac{c_{j 2}}{{(X - a_{j})}^{2}} + \dots + \frac{c_{j l_{j}}}{{(X - a_{j})}^{l_{j}}} \\ + \frac{m_{11} X + b_{11}}{X^{2} + p_{1} X + q_{1}} + \frac{m_{12} X + b_{12}}{(X^{2} + p_{1} X + q_{1})^{2}} + \dots + \frac{m_{1 n_{1}} X + b_{1 n_{1}}}{(X^{2} + p_{1} X + q_{1})^{n_{1}}} \\ + \dots \\ + \frac{m_{k 1} X + b_{k 1}}{X^{2} + p_{k} X + q_{k}} + \frac{m_{k 2} X + b_{k 2}}{(X^{2} + p_{k} X + q_{k})^{2}} + \dots + \frac{m_{k n_{k}} X + b_{k n_{k}}}{(X^{2} + p_{k} X + q_{k})^{n_{k}}} \end{array}$

t being a suitable polynomial such that $t = 0$ if $grad r < grad s$ and $grad t = grad r - grad s$ otherweise.

Both theorems, especially the fundamental theorem, are pure existence theorems! And it is this fact that causes actual problems when it comes to application: Apart from simple and clear cut cases it is nearly impossible to get the necessary denominator's decomposition.

Example: We find a partial decomposition for

f = \frac{2 X^{5} - 4 X^{4} + 10 X^{3} - 17 X^{2} + 6 X - 3}{X^{4} - 2 X^{3} + 3 X^{2} - 4 X + 2}

First we (sequentially) discover that 1 is a double zero of the denominator so that a twofold polynomial division yields the decomposition

$X^{4} - 2 X^{3} + 3 X^{2} - 4 X + 2 = {(X - 1)}^{2} \cdot (X^{2} + 2)$
and a further polynomial division then provides the identity

$f = 2 X + \frac{4 X^{3} - 9 X^{2} + 2 X - 3}{{(X - 1)}^{2} \cdot (X^{2} + 2)}$
Finally, comparing the coefficients in the ansatz

$\begin{array}{l} \frac{4 X^{3} - 9 X^{2} + 2 X - 3}{{(X - 1)}^{2} \cdot (X^{2} + 2)} \\ = & \frac{c_{11}}{X - 1} + \frac{c_{12}}{{(X - 1)}^{2}} + \frac{m_{11} X + b_{11}}{X^{2} + 2} \\ = & \frac{c_{11} (X - 1) (X^{2} + 2) + c_{12} (X^{2} + 2) + (m_{11} X + b_{11}) {(X - 1)}^{2}}{{(X - 1)}^{2} (X^{2} + 2)} \\ = & \frac{(c_{11} + m_{11}) X^{3} + (- c_{11} + c_{12} - 2 m_{11} + b_{11}) X^{2} + (2 c_{11} + m_{11} - 2 b_{11}) X - 2 c_{11} + 2 c_{12} + b_{11}}{{(X - 1)}^{2} (X^{2} + 2)} \end{array}$

yields the following linear system of equations

$\begin{array}{l} c_{11} + m_{11} & = 4 & \Leftrightarrow & c_{11} & = 0 \\ - c_{11} + c_{12} - 2 m_{11} + b_{11} & = - 9 & c_{12} & = - 2 \\ 2 c_{11} + m_{11} - 2 b_{11} & = 2 & m_{11} & = 4 \\ - 2 c_{11} + 2 c_{12} + b_{11} & = - 3 & b_{11} & = 1 \end{array}$

f thus decomposes like this

$f = 2 X - \frac{2}{{(X - 1)}^{2}} + \frac{4 X + 1}{X^{2} + 2}$ [2]

With a partial decomposition of f at hand, a primitive of f is easily found if we know primitives for each addend. This reduces our problem to only two types of rational functions, namely ( $k \in ℕ$ )

$\frac{c}{{(X - a)}^{k}}$
$\frac{m X + b}{(X^{2} + p X + q)^{k}}$ where $D = \frac{p^{2}}{4} - q < 0$

Quotients of the first kind are easy to deal with, except for the case $k = 1$ which needs the natural logarithm ln
i

from chapter 8.

Proposition:

$c \cdot \ln \| X - a \|$ is a primitive function of $\frac{c}{X - a}$ .	[8.0.1]
If $k > 1$ then $\frac{1}{1 - k} \cdot \frac{c}{{(X - a)}^{k - 1}}$ is a primitive function of $\frac{c}{{(X - a)}^{k}}$ .	[8.0.2]

Proof:

1. ► In [8.7.1] we introduce ln as a primitive of the reciprocal function on $ℝ^{> 0}$ . ln is thus differentiable and

\ln^{'} (x) = \frac{1}{x}

for all

x > 0

Using the chain rule [7.7.8] and the derivative of the absolute value function [7.4.3] we see that $c \cdot \ln | X - a |$ is differentiable and that

(c \cdot \ln | X - a |)^{'} = c \cdot \frac{1}{| X - a |} \cdot | X - a |^{'} = c \cdot \frac{1}{| X - a |} \cdot \frac{| X - a |}{X - a} = \frac{c}{X - a}

is its derivative.

2. ► $\frac{1}{1 - k} \cdot \frac{c}{{(X - a)}^{k - 1}}$ is essentially a power of X, thus differentiable. We use the power rule
i

$(X^{n})^{'} = n X^{n - 1}$ for all $n \in ℤ$

to calculate the derivative:

(\frac{1}{1 - k} \cdot \frac{c}{{(X - a)}^{k - 1}})^{'} = \frac{1}{1 - k} \cdot (- (k - 1)) \cdot \frac{c}{{(X - a)}^{k}} = \frac{c}{{(X - a)}^{k}}

Examples are quite straight usually:

$4 \cdot \ln | X - 7 |$ is a primitive of $\frac{4}{X - 7}$ .
$- \frac{2}{X - 1}$ is a primitive of $\frac{2}{{(X - 1)}^{2}}$ .

Dealing with quotients of the second kind is a bit catchier. But as we always have the decomposition

\frac{m X + b}{(X^{2} + p X + q)^{k}} = \frac{m}{2} \cdot \frac{2 X + p}{(X^{2} + p X + q)^{k}} + \frac{b - \frac{m}{2} p}{(X^{2} + p X + q)^{k}}

[3]

we may confine ourselves only to the cases $\frac{2 X + p}{(X^{2} + p X + q)^{k}}$ and $\frac{c}{(X^{2} + p X + q)^{k}}$ . The first one is easy.

Proposition:

$\ln (X^{2} + p X + q)$ is a primitive function of $\frac{2 X + p}{X^{2} + p X + q}$ .	[8.0.3]
If $k > 1$ then $\frac{1}{1 - k} \cdot \frac{1}{(X^{2} + p X + q)^{k - 1}}$ is a primitive function of $\frac{2 X + p}{(X^{2} + p X + q)^{k}}$ .	[8.0.4]

Proof: Both assertions are easily proved by using the chain rule. In 1. we note that, due to the premise, the values of $X^{2} + p X + q$ are always $> 0$ . The domain of $\ln (X^{2} + p X + q)$ is thus the whole of $ℝ$ .

As an example we see that

$\ln (X^{2} + 2)$ is a primitive function of $\frac{2 X}{X^{2} + 2}$ .
$- \frac{1}{3} \cdot \frac{1}{(X^{2} - 3 X + 5)^{3}}$ is a primitive function of $\frac{2 X - 3}{(X^{2} - 3 X + 5)^{4}}$ .

The remaining case is actually quite cumbersome. We recall that the discriminant $D = \frac{p^{2}}{4} - q$ is negative, thus $- D$ is positive. At first we show that it is sufficient to consider only denominators like $(X^{2} + 1)^{k}$ .

Proposition: If g is a primitive of $\frac{c}{(X^{2} + 1)^{k}}$ then

\sqrt{- D^{1 - 2 k}} \cdot g \circ \frac{X + \frac{p}{2}}{\sqrt{- D}}

[8.0.5]

is a primitive function of $\frac{c}{(X^{2} + p X + q)^{k}}$ .

Proof: The chain rule guarantees the differentiability of the function in [8.0.5] and provides the following derivative:

\begin{array}{l} (\sqrt{- D^{1 - 2 k}} \cdot g \circ \frac{X + \frac{p}{2}}{\sqrt{- D}})^{'} & = \sqrt{{(- D)}^{1 - 2 k}} \cdot (g^{'} \circ \frac{X + \frac{p}{2}}{\sqrt{- D}}) \cdot \frac{1}{\sqrt{- D}} \\ = \sqrt{{(- D)}^{- 2 k}} \cdot \frac{c}{(X^{2} + 1)^{k}} \circ \frac{X + \frac{p}{2}}{\sqrt{- D}} \\ = {(- D)}^{- k} \cdot \frac{c}{{(\frac{{(X + \frac{p}{2})}^{2}}{- D} + 1)}^{k}} \\ = {(- D)}^{- k} \cdot \frac{c \cdot {(- D)}^{k}}{{(X^{2} + p X + \frac{p^{2}}{4} - D)}^{k}} \\ = \frac{c}{(X^{2} + p X + q)^{k}} \end{array}

Primitives of $(X^{2} + 1)^{k}$ are eventually constructed by recursion. For the base step we need the invers tangent,

\arctan : ℝ \to] - \frac{π}{2}, \frac{π}{2} [

the inverse of $\tan |] - \frac{π}{2}, \frac{π}{2} [$
i

The restriction $\tan |] - \frac{π}{2}, \frac{π}{2} [$ is bijective as it is

injective due to [7.9.6] because

$\tan^{'} (x) = \frac{1}{\cos^{2} (x)} \neq 0$ for all $x \in] - \frac{π}{2}, \frac{π}{2} [$ .

surjective according to a consequence of the intermediate value theorem [6.6.2] based on the limits

$\lim_{\begin{array}{c} x \to \pm \frac{π}{2} \\ x \in] - \frac{π}{2}, \frac{π}{2} [ \end{array}} \tan x = \pm \infty$

.

Proposition:

$c \cdot \arctan$ is a primitive function of $\frac{c}{X^{2} + 1}$ .	[8.0.6]
If g is a primitive of $\frac{c}{(X^{2} + 1)^{k}}$ then
$\frac{1}{2 k} \cdot (\frac{c X}{(X^{2} + 1)^{k}} + (2 k - 1) \cdot g)$	[8.0.7]

is a primitive function of $\frac{c}{(X^{2} + 1)^{k + 1}}$ .

Proof:

1. ► As $\tan^{'} (x) = \frac{1}{\cos^{2} (x)} \neq 0$ for all $x \in] - \frac{π}{2}, \frac{π}{2} [$ , arctan is differentiable due to [7.5.4] with

\arctan^{'} (x) = \frac{1}{\tan^{'} (\arctan x)} = \cos^{2} (\arctan x) \underset{[+]}{=} \frac{1}{\tan^{2} (\arctan x) + 1} = \frac{1}{x^{2} + 1}

Note that [+] is valid according to the identity $\cos^{2} = \frac{1}{\tan^{2} + 1}$ .

2. ► The function in [8.0.7] is differentiable due to the quotient rule [7.7.7]. Its derivative calculates to:

\begin{array}{l} \frac{1}{2 k} \cdot (\frac{c X}{(X^{2} + 1)^{k}} + (2 k - 1) \cdot g)^{'} & = \frac{1}{2 k} \cdot (\frac{c (X^{2} + 1)^{k} - c X \cdot k (X^{2} + 1)^{k - 1} \cdot 2 X}{(X^{2} + 1)^{2 k}} + (2 k - 1) \cdot \frac{c}{(X^{2} + 1)^{k}} \\ = \frac{c}{2 k} \cdot \frac{X^{2} + 1 - 2 k X^{2} + (2 k - 1) \cdot (X^{2} + 1)}{(X^{2} + 1)^{k + 1}} \\ = \frac{c}{2 k} \cdot \frac{2 k}{(X^{2} + 1)^{k + 1}} \\ = \frac{c}{(X^{2} + 1)^{k + 1}} \end{array}

An example will show how this procedure works.

Example:

It takes three steps to construct a primitive function of $\frac{2}{(X^{2} + 1)^{3}}$ recursively by [8.0.7]:

$1 .) 2 \cdot \arctan$ is a primitive of $\frac{2}{(X^{2} + 1)}$ .

$\underset{k = 1}{2 .)} \frac{1}{2} \cdot (\frac{2 X}{X^{2} + 1} + 2 \cdot \arctan) = \frac{X}{X^{2} + 1} + \arctan$ is a primitive of $\frac{2}{(X^{2} + 1)^{2}}$ .

$\underset{k = 2}{3 .)} \frac{1}{4} \cdot (\frac{2 X}{(X^{2} + 1)^{2}} + 3 \cdot (\frac{X}{X^{2} + 1} + \arctan)) = \frac{1}{2} \cdot \frac{X}{(X^{2} + 1)^{2}} + \frac{3}{4} \cdot \frac{X}{X^{2} + 1} + \frac{3}{4} \cdot \arctan$
is a primitive of $\frac{2}{(X^{2} + 1)^{3}}$ .
To get a primitive of $\frac{2}{(X^{2} + 6 X + 13)^{2}}$ we start by calculating the discriminant $D = - 4$ from the data $p = 6$ and $q = 13$ . With step two of the example above and with [8.0.5] we then find

$\begin{array}{l} \sqrt{- D^{1 - 2 k}} \cdot g \circ \frac{X + \frac{p}{2}}{\sqrt{- D}} & = \frac{1}{\sqrt{64}} \cdot (\frac{X}{X^{2} + 1} + \arctan) \circ \frac{X + 3}{2} \\ = \frac{1}{8} \cdot (\frac{\frac{X + 3}{2}}{\frac{{(X + 3)}^{2} + 4}{4}} + \arctan \frac{X + 3}{2}) \\ = \frac{1}{4} \cdot \frac{X + 3}{X^{2} + 6 X + 13} + \frac{1}{8} \cdot \arctan \frac{X + 3}{2} \end{array}$

as a primitive function.
Finally, to get a primitive of $\frac{3 X + 11}{(X^{2} + 6 X + 13)^{2}}$ we first consider the decomposition

$\frac{3 X + 11}{(X^{2} + 6 X + 13)^{2}} = \frac{3}{2} \cdot \frac{2 X + 6}{(X^{2} + 6 X + 13)^{2}} + \frac{2}{(X^{2} + 6 X + 13)^{2}}$

according to [3]. From the previous result and from [8.0.4] we now find

$\begin{array}{l} - \frac{3}{2} \cdot \frac{1}{X^{2} + 6 X + 13} + \frac{1}{4} \cdot \frac{X + 3}{X^{2} + 6 X + 13} + \frac{1}{8} \cdot \arctan \frac{X + 3}{2} \\ = & \frac{1}{4} \cdot \frac{X - 3}{X^{2} + 6 X + 13} + \frac{1}{8} \cdot \arctan \frac{X + 3}{2} \end{array}$

as a primitive.

We now return to our initial example

f = \frac{2 X^{5} - 4 X^{4} + 10 X^{3} - 17 X^{2} + 6 X - 3}{X^{4} - 2 X^{3} + 3 X^{2} - 4 X + 2}

In [2] we showed that

2 X - \frac{2}{{(X - 1)}^{2}} + \frac{4 X + 1}{X^{2} + 2} = 2 X - \frac{2}{{(X - 1)}^{2}} + 2 \cdot \frac{2 X}{X^{2} + 2} + \frac{1}{X^{2} + 2}

is its partial fraction decomposition. Thus f is completely decomposed in processable basic typs and a primitive of f is simply gained by adding their primitive functions:

X^{2} + \frac{2}{X - 1} + 2 \cdot \ln (X^{2} + 2) + \frac{1}{\sqrt{2}} \cdot \arctan \frac{X}{\sqrt{2}}