
We discuss an approach that allows us to integrate rational functions.

### Basics of polynomial and rational functions

In this course, we are attempting to learn to work with as many functions as possible. A basic class of functions are polynomial functions:

Which of the following are polynomial functions?
$f(x) = 0$ $f(x) = 3x+1$ $f(x) = x^{1/2}-x +8$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = (x+1)(x-1)+e^x - e^x$ $f(x) = \frac {x^2 - 3x + 2}{x-2}$ $f(x) = x^7-32x^6-\pi x^3+45/84$

As we will see, there is a fact about polynomials that is of critical importance for this section:

Polynomials are equal as functions if and only if their coefficients are equal.

Given two polynomials equal as functions: What are $a_0$, $a_1$, $a_2$, $a_3$, $a_4$, $a_5$?
• $a_0 = \answer {-5}$
• $a_1 = \answer {0}$
• $a_2 = \answer {-1}$
• $a_3 = \answer {0}$
• $a_4 = \answer {-24}$
• $a_5 = \answer {6}$

In the world of mathematics, polynomials are a generalization of “integers,” and rational numbers are fractions of integers. This brings us to our next definition:

Which of the following are rational functions?
$f(x) = 0$ $f(x) = \frac {3x+1}{x^2-4x+5}$ $f(x)=e^x$ $f(x)=\frac {\sin (x)}{\cos (x)}$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = x^{1/2}-x +8$ $f(x)=\frac {\sqrt {x}}{x^3-x}$

### Denominators with distinct linear factors

We are already skilled at working with polynomials, we can differentiate and integrate any polynomial function. Being able to integrate any rational function is the next logical step in our (rather ambitious) quest to integrate all functions. Let’s dig right in with an example.

What we have seen is part of a general technique of integration called “partial fractions” that, in principle, allows us to integrate any rational function.

#### The general technique for distinct linear factors

Suppose you wish to compute where $p$ and $q$ are both polynomial functions, the degree of $p$ is less than the degree of $q$, and $q$ factors into $n$ distinct linear factors: then we can always write The right-hand side of the equation above is easy to antidifferentiate, as we can integrate it term-by-term and hence

### Denominators with repeated linear factors

Here we work as we did before, except we add an extra variable for each of the repeated factors. Let’s do an example.

#### The general technique for repeated linear factors

Suppose you wish to compute where $p$ and $q$ are both polynomial functions, the degree of $p$ is less than the degree of $q$, and $q$ factors into $n$ repeated linear factors: then we can always write The right-hand side of the equation above is easy to antidifferentiate, as we can integrate it term-by-term and

hence

### Denominators with distinct irreducible quadratic factors

Here is a fact about polynomials:

Remember, a root is where a polynomial is zero. The theorem above is a deep fact of mathematics. The great mathematician Gauss proved the theorem in 1799 for his doctoral thesis. This fact can be used to show the following:

Every polynomial function will factor as a product of linear terms and irreducible quadratic terms over the real numbers.

So now let’s work an example where the denominator of our rational function has distinct quadratic factors.

#### The general technique for distinct quadratic factors

Suppose you wish to compute where $p$ and $q$ are both polynomial functions, the degree of $p$ is less than the degree of $q$, and $q$ factors into $n$ distinct irreducible quadratic factors: then we can always write The right-hand side of the equation can be antidifferentiated, though it is not always “easy.”

### Denominators with repeated quadratic factors

For completeness sake, we will work a problem with repeated quadratic factors.

#### The general technique for repeated quadratic factors

Suppose you wish to compute where $p$ and $q$ are both polynomial functions, the degree of $p$ is less than the degree of $q$, and $q$ factors into $n$ distinct irreducible quadratic factors: then we can always write The right-hand side of the equation can be antidifferentiated, though it is not always “easy.”

### Reducing rational functions

When computing all of the techniques above rely on the fact that the degree of $p$ is less than the degree of $q$. What if this is not the case? Use long-division.

As with many other problems in calculus, it is important to remember that one is not expected to “see” the final answer immediately after seeing the problem. Rather, given the initial problem, we break it down into smaller problems that are easier to solve. The final answer is a combination of the answers of the smaller problems.