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Section 7.4 Partial Fractions

So far we have not learned how to integrate general rational functions.

Definition 7.4.1.
A rational function is in the form \(f(x) = \dfrac{\text{polynomial}}{\text{polynomial}}\text{.}\)

In order to do this we will use partial fractions. An example of how partial fractions work is as follows:

\begin{align*} \dfrac{1}{x} - \dfrac{1}{x+1} \amp = \dfrac{1}{x}\cdot\dfrac{x+1}{x+1} - \dfrac{1}{x+1}\cdot\dfrac{x}{x} \\ \amp = \dfrac{x+1}{x(x+1)} - \dfrac{x}{x(x+1)} \\ \amp = \dfrac{x+1-x}{x(x+1)}\\ \amp = \dfrac{1}{x(x+1)}\\ \amp = \dfrac{1}{x^2+x} \end{align*}

The only difference in how we use it for integrals is that the process is reversed:

\begin{align*} \int \dfrac{1}{x^2 + x} dx \amp = \int \dfrac{1}{x} - \dfrac{1}{x+1} dx \\ \amp = \int \dfrac{1}{x} dx - \int \dfrac{1}{x+1} dx \\ \amp = \ln | x | - \ln | x - 1 | + c \end{align*}

Subsection 7.4.1 General Approach

Let's find the partion fraction decomposition of the following:

\begin{equation*} \dfrac{5x-10}{x^2-3x-4} = \dfrac{5x-10}{(x-4)(x+1)} = \dfrac{A}{x-4} + \dfrac{B}{x+1} \end{equation*}

How do we find A and B? Multiply both sides by \((x-4)(x+1)\) to get:

\begin{align*} (x-4)(x+1) \cdot \dfrac{5x-10}{(x-4)(x+1)} \amp = \left( \dfrac{A}{x-4} + \dfrac{B}{x+1} \right) \cdot (x-4)(x+1) \\ 5x - 10 \amp = A(x+1) + B(x-4) \end{align*}

We want this to be true for all x:

Let \(x = 4\text{:}\)

\begin{align*} 5(4) - 10 \amp = A (x - 4) + B (x + 1) \\ 10 \amp = 5A\\ A \amp = 2 \end{align*}

Let \(x = -1\text{:}\)

\begin{align*} 5(-1) - 10 \amp = A (-1 + 1) + B (-1 - 4)\\ -15 \amp = -5 B\\ B = 3 \end{align*}
So,

\begin{equation*} \dfrac{5x-10}{x^2-3x-4} = \dfrac{2}{x-4} + \dfrac{3}{x+1} \end{equation*}

The previous example is the simplest and most general case. There are 3 types of problems that reequire differing methods:

  1. repeated factors

  2. irreducible quadratic factors

  3. degree of numberator \(\ge\) denominator

Subsection 7.4.2 Repeated Factors

One of the issues that we come across when evaluating partial fractions is when there are repeated factors in the denominator. An example of this would be \(\dfrac{x+3}{(x+2)^2}\text{.}\) If we simplify this it would be \(\dfrac{x+3}{(x+8)(x+8)}\text{.}\) When we have this issue we cannot follow the normal procedure:

\begin{equation*} \dfrac{x+3}{(x+8)(x+8)} \neq \dfrac{A}{x+8} + \dfrac{B}{x+8} \end{equation*}

In order to evaluate this correctly, we do the following:

\begin{align*} \dfrac{x+3}{(x+8)(x+8)} \amp = \dfrac{A}{x+8} + \dfrac{B}{(x+8)^2} \\ 1(x) + 3 \amp = A(x+8) + B\\ \amp = Ax + 8A + B \end{align*}

* multiply both sides by \((x+8)(x+8)\)

\begin{gather*} A = 1 \end{gather*}
\begin{align*} 8A + B \amp = 3\\ 8(1) + B \amp = 3\\ B \amp = -5 \end{align*}
So, \(\dfrac{x+3}{(x+8)^2} = \dfrac{1}{x+8} + \dfrac{-5}{(x+5)^2}\)

If there are highter exponents for the denominator, you get something that is in the form of the following:

\begin{equation*} \dfrac{5x+3}{x(x+5)^3} = \dfrac{A}{x} + \dfrac{B}{x+5} + \dfrac{C}{(x+5)^2} + \dfrac{D}{(x+5)^3} \end{equation*}

Subsection 7.4.3 Irreducable Quadratic Factors

Another type of partial fractions that will give us some trouble is one that contains an irreducable quadratic factor.

Definition 7.4.2.

An irreducable factor is one that has no real zeros; it cannot factor without using imaginary numbers.

Again we cannot follow the normal procedure:

\begin{equation*} \dfrac{7x^2+5x+18}{x(x^2+9)} \neq \dfrac{A}{x} + \dfrac{B}{x^2 + 9} \end{equation*}

Recall: \((x^2 + 9)\) would simplify to \((x + 3i)(x - 3i)\)

Note 7.4.3.

There are two ways to determine if a function has an irreducable factor:

  • using a graph

    • reducable: touches the x-axis

    • irreducable: never touches the x-axis

  • formula: if \(b^2 - 4ac \geq 0\text{,}\) then reducable, otherwise irreducable.

Also, keep in mind, every polynomial function with real coefficients can be factored into linear and/or irreducable quadratic functions.

We have to follow the general steps below to deal with irreducables:
\begin{align*} \dfrac{7x^2+5x+18}{x(x^2+9)} \amp = \dfrac{A}{x} + \dfrac{Bx+C}{x^2 + 9} \\ 7x^2+5x+18 \amp = A(x^2+9) + x(Bx + C) \\ \amp = Ax^2 + 9A + Bx^2 + Cx\\ \amp = (A + B)x^2 + Cx + 9A \end{align*}

* multiply both sides by \(x(x^2+9)\)

\begin{equation*} C = 8 \end{equation*}
\begin{align*} 9A \amp = 18 \\ A \amp = 2 \end{align*}
\begin{align*} A + B \amp = 7 \\ 2 + B \amp = 7 \\ B \amp = 5 \end{align*}

So, \(\dfrac{7x^2+5x+18}{x(x^2+9)} = \dfrac{2}{x} + \dfrac{5x+8}{x^2+9} \)

If we mix the last two methods we would get something of the form:

\begin{equation*} \dfrac{9x^4 - 2x - 13}{(x-3)(x^2+7)(x^2+25)^2} = \dfrac{A}{x-3} + \dfrac{Bx+C}{x^2+7} \dfrac{Dx+E}{x^2+25} + \dfrac{Fx+G}{(x^2+25)^2} \end{equation*}

Subsection 7.4.4 Degree

The final exception is when the degree of the numberator is greater than or equal to the degree of the denominator. For this type of problem we will have to use polynomial long division where the numerator is divided by the denominator.

Example: Divide \(x^4+3x^2-8\) by \(x^2-2x+1\)

Table 7.4.4. Polynomial Division
\(\) \(\) \(x^2 \) \(+ 2x \) \(+ 6\) \(+ \dfrac{10x-14}{x^2-2x+1}\)
\(x^2-2x+1\) \(x^4\) \(+0x^3\) \(+3x^2\) \(+0x\) \(-8\)
\(\) \(-(x^4 \) \(-2x^3 \) \(+x^2) \) \(\) \(\) \(\)
\(\) \(\) \(2x^3 \) \(+2x^2 \) \(+0x \) \(\) \(\)
\(\) \(\) \(-(2x^3 \) \(-4x^2 \) \(+2x) \) \(\) \(\)
\(\) \(\) \(\) \(6x^2 \) \(-2x \) \(-8 \) \(\)
\(\) \(\) \(\) \(-(6x^2 \) \(-12x \) \(+6) \) \(\)
\(\) \(\) \(\) \(\) \(10x \) \(-14 \) \(\)

\begin{align*} \text{So, } \int \dfrac{x^4+3x^2-8}{x^2-2x+1} dx \amp= \int x^2+2x+6 + \dfrac{10x-14}{x^2-2x+1} dx \\ \amp = \int x^2+2x+6 + \dfrac{10x-14}{(x-1)^2} dx \\ \amp = \int x^2+2x+6 + \dfrac{10}{x-1} - \dfrac{14}{(x-1)^2} dx \\ \amp \vdots \end{align*}

We would then just follow the same steps as with any of the previous examples.