Rational functions

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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:

A polynomial function in the variable $x$ is a function which can be written in the form $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ where the $a_k$ ’s are all constants (called the coefficients) and $n$ is a whole number (called the degree when $n\ne 0$ ). The domain of a polynomial function is $(-\infty ,\infty )$ .

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: $6x^5+a_4 x^4 -x^2 + a_0 = a_5 x^5 - 24 x^4 + a_3 x^3 + a_2 x^2 - 5$ 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:

A rational function in the variable $x$ is a function the form $f(x) = \frac {p(x)}{q(x)}$ where $p$ and $q$ are polynomial functions. The domain of a rational function is all real numbers except for where the denominator is equal to zero.

Which of the following are rational functions?

$f(x) = 0$ $f(x) = \frac {3x+1}{x^2-4x+5}$ $f(x)=e^x$ $f(x)=\frac {\sin (x)}{\cos (x)}$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = x^{1/2}-x +8$ $f(x)=\frac {\sqrt {x}}{x^3-x}$

All polynomials can be thought of as rational functions.

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.

Compute: $\int \frac {1}{x^2-1} \d x$

We will suppose that there are numbers $A$ and $B$ such that $\frac {A}{x-1} + \frac {B}{x+1} = \frac {1}{x^2-1}.$ Clearing denominators, we find $A\answer [given]{(x+1)} + B\answer [given]{(x-1)} = 1.$ Expanding the left-hand side, we find a polynomial equal (as a function!) to the constant polynomial $1$ $(A+ B)x + (A-B) = 1$ Since polynomials are equal as functions if and only if their coefficients are equal, we may rewrite this as two equations: $\begin{align*} \answer[given]{A+B} &= 0 &\text{(the coefficients for $x$)}\\ \answer[given]{A-B} &= 1 &\text{(the coefficients for the constant)} \end{align*}$ Solving these three equations for $A$ and $B$ we find

$A = \answer [given]{1/2}$ ,
$B = \answer [given]{-1/2}$ .

From this we can now rewrite our integral as

$\begin{align*} \int \frac{1}{x^2-1} \d x &= \int \frac{1/2}{x-1} + \frac{-1/2}{x+1} \d x\\ &= (1/2)\ln|x-1| + (-1/2)\ln|x+1|+ K. \end{align*}$

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 $\int \frac {p(x)}{q(x)} \d x$ 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: $q(x) = (x-r_1) (x-r_2) \cdots (x-r_n),$ then we can always write $\frac {p(x)}{q(x)} = \frac {A_1}{(x-r_1)} + \frac {A_2}{(x-r_2)} + \cdots + \frac {A_n}{(x-r_n)}.$ The right-hand side of the equation above is easy to antidifferentiate, as we can integrate it term-by-term and $\int \frac {A_k}{(x-r_k)} \d x = A \ln (x-r_k) + K,$ hence $\int \frac {p(x)}{q(x)} \d x = \sum _{k=1}^n A_k \ln |x-r_k| +K.$

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.

Compute: $\int \frac {1}{(x-1)(x+2)^2} \d x$

We will suppose that there are numbers $A$ , $B$ , and $C$ such that $\frac {A}{x-1} + \frac {B}{x+2} + \frac {C}{(x+2)^2} = \frac {1}{(x-1)(x+2)^2}$ Clearing denominators, we find $A\answer [given]{(x+2)^2} + B\answer [given]{(x-1)(x+2)} + C\answer [given]{(x-1)} = 1.$ Expanding the left-hand side, we find a polynomial equal (as a function!) to the constant polynomial $1$ $(A+B)x^2 + (4A+B+C)x + (4A-2B-C) = 1$ Since polynomials are equal as functions if and only if their coefficients are equal, we may rewrite this as three equations: $\begin{align*} \answer[given]{A+B} &= 0 &\text{(the coefficients for $x^2$)}\\ \answer[given]{4A+B+C} &= 0 &\text{(the coefficients for $x$)}\\ \answer[given]{4A-2B-C} &= 1 &\text{(the coefficients for the constant)} \end{align*}$ Solving these three equations for $A$ , $B$ and $C$ we find

$A = \answer [given]{1/9}$ ,
$B = \answer [given]{-1/9}$ ,
$C = \answer [given]{-1/3}$ .

From this we can now rewrite our integral as

$\begin{align*} \int&\frac{1}{(x-1)(x+2)^2}\d x= \int \frac{1/9}{x-1}+ \frac{-1/9}{x+2} + \frac{-1/3}{(x+2)^2}\d x\\ &= \int \frac{1/9}{x-1}\d x + \int \frac{-1/9}{x+2}\d x + \int \frac{-1/3}{(x+2)^2}\d x,\\ &= \frac{1}{9}\ln|x-1| -\frac{1}{9}\ln|x+2| +\frac{1}{3(x+2)}+K. \end{align*}$

The general technique for repeated linear factors

Suppose you wish to compute $\int \frac {p(x)}{q(x)} \d x$ 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: $q(x) = (x-r)^n,$ then we can always write $\frac {p(x)}{q(x)} = \frac {A_1}{(x-r)} + \frac {A_2}{(x-r)^2} + \cdots + \frac {A_n}{(x-r)^n}.$ The right-hand side of the equation above is easy to antidifferentiate, as we can integrate it term-by-term and

$\begin{align*} \int \frac{A_1}{(x-r)} \d x &= A_1 \ln|x-r| + K,\\ \int \frac{A_k}{(x-r)^k} \d x &= \frac{A_k}{(1-k)(x-r)^{k-1}}+ K, &\text{(if $k>1$)} \end{align*}$ hence $\int \frac {p(x)}{q(x)} \d x = A_1 \ln |x-r| + \sum _{k=2}^n \frac {A_k}{(1-k)(x-r)^{k-1}} + K.$

Denominators with distinct irreducible quadratic factors

Here is a fact about polynomials:

The Fundamental Theorem of Algebra Every polynomial of the form $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ where the $a_i$ ’s are real (or even complex!) numbers and $a_n \ne 0$ has exactly $n$ (possibly repeated) complex roots.

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.

Compute: $\int \frac {7x^2+31x+54}{(x+1)(x^2+6x+11)}\d x$

We will suppose that there are numbers $A$ , $B$ , and $C$ such that $\frac {A}{x+1} + \frac {Bx+C}{x^2+6x+11} = \frac {7x^2+31x+54}{(x+1)(x^2+6x+11)}.$ Clearing denominators, we find $\left (\answer [given]{x^2+6x+11}\right )A + (Bx+C)\left (\answer [given]{x+1}\right ) = 7x^2+31x+54.$ Expanding the left-hand side, we find a polynomial equal (as a function!) to the polynomial $7x^2+31x+54$ $(A+B)x^2 + (6A+B+C)x + (11A+C) = 7x^2+31x+54.$ Since polynomials are equal as functions if and only if their coefficients are equal, we may rewrite this as three equations: $\begin{align*} \answer[given]{A+B} &= 7 &\text{(the coefficients for $x^2$)}\\ \answer[given]{6A+B+C} &= 31 &\text{(the coefficients for $x$)}\\ \answer[given]{11A+C} &= 54 &\text{(the coefficients for the constant)} \end{align*}$ Solving these three equations for $A$ , $B$ and $C$ we find

$A = \answer [given]{5}$ ,
$B = \answer [given]{2}$ ,
$C = \answer [given]{-1}$ .

From this we can now rewrite our integral as $\int \frac {7x^2+31x+54}{(x+1)(x^2+6x+11)}\d x = \int \frac {5}{x+1} + \frac {2x-1}{x^2+6x+11} \d x\\$ The first term of this new integrand is easy to evaluate. We find $\int \frac {5}{x+1} \d x = 5\ln |x+1|+K.$ The second term is not hard, but takes several steps and uses substitution techniques.

The integrand $\frac {2x-1}{x^2+6x+11}$ has a quadratic in the denominator and a linear term in the numerator. This leads us to try substitution. Let

$\begin{align*} g &= x^2+6x+11,\\ \d g &= (2x+6)\d x. \end{align*}$ However, the numerator is $2x-1$ , not $2x+6$ ! We can bypass this difficulty by adding “ $0$ ” in the form of “ $7-7$ .” $\begin{align*} \frac{2x-1}{x^2+6x+11} &= \frac{2x-1+7-7}{x^2+6x+11} \\ &= \frac{2x+6}{x^2+6x+11} - \frac{7}{x^2+6x+11}. \end{align*}$ We can now integrate the first term with substitution, leading to $\int \frac {2x+6}{x^2+6x+11} \d x = \ln |x^2+6x+11| + K.$ The final term can be integrated using arctangent. First, complete the square in the denominator: $\frac {7}{x^2+6x+11} = \frac {7}{(x+3)^2+2}.$ then use a substitution of $g = x+3$ to find $\int \frac {7}{x^2+6x+11}\d x = \frac {7}{\sqrt {2}}\arctan \left (\frac {x+3}{\sqrt {2}}\right )+K.$ Let’s start at the beginning and put all of the steps together. $\begin{align*} \int&\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\d x \\ &= \int\left(\frac{5}{x+1} + \frac{2x-1}{x^2+6x+11}\right)\d x \end{align*}$ breaking this integral up we find $\begin{align*} = \int\frac{5}{x+1}\d x &+ \int\frac{2x+6}{x^2+6x+11}\d x\\ &-\int\frac{7}{x^2+6x+11}\d x \end{align*}$ and antidifferentiating we find $= 5\ln |x+1|+ \ln |x^2+6x+11|-\frac {7}{\sqrt {2}}\arctan \left (\frac {x+3}{\sqrt {2}}\right )+K.$

The general technique for distinct quadratic factors

Suppose you wish to compute $\int \frac {p(x)}{q(x)} \d x$ 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: $q(x) = (a_1x^2 + b_1 x + c_1) (a_2x^2 + b_2 x + c_2)\cdots (a_nx^2 + b_n x + c_n)$ then we can always write $\frac {p(x)}{q(x)} = \frac {A_1x+B_1}{a_1x^2 + b_1 x + c_1} + \cdots + \frac {A_nx+B_n}{a_nx^2 + b_n x + c_n}.$ 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.

Compute $\int \frac {1}{x^5 + 2x^3 + x}\d x$

Start by factoring the denominator $x^5 + 2x^3 + x = x(x^2+1)^2.$ We will suppose that there are numbers $A$ , $B$ , $C$ , $D$ , and $E$ such that $\frac {A}{x} + \frac {Bx+C}{x^2+1} + \frac {Dx+E}{(x^2+1)^2} = \frac {1}{x(x^2+1)^2}.$ Clearing denominators, we find $A\answer [given]{(x^2+1)^2} + (Bx+C)\answer [given]{(x^3+x)}+ (Dx+E)\answer [given]{(x)} = 1$ Expanding the left-hand side, we find a polynomial equal (as a function!) to the constant $1$ . $(A+B)x^4 + Cx^3 + (2A+B+D)x^2 + (C+E)x + A = 1.$ Since polynomials are equal as functions if and only if their coefficients are equal, we may rewrite this as five equations: $\begin{align*} \answer[given]{A+B} &= 0 &\text{(the coefficients for $x^4$)}\\ \answer[given]{C} &= 0 &\text{(the coefficients for $x^3$)}\\ \answer[given]{2A+B+D} &= 0 &\text{(the coefficients for $x^2$)}\\ \answer[given]{C+E} &= 0 &\text{(the coefficients for $x$)}\\ \answer[given]{A} &= 1 &\text{(the coefficients for the constant)} \end{align*}$ Solving these three equations for $A$ , $B$ , $C$ , $D$ , and $E$ we find

$A = \answer [given]{1}$ ,
$B = \answer [given]{-1}$ ,
$C = \answer [given]{0}$ ,
$D = \answer [given]{-1}$ ,
$E = \answer [given]{0}$ .

From this we can now rewrite our integral as $\int \frac {1}{x^5 + 2x^3 + x}\d x =\int \frac {1}{x} + \frac {-x}{x^2+1} + \frac {-x}{(x^2+1)^2}\d x$ Each term of this new integrand is easy to evaluate, write

$\begin{align*} \int \frac{1}{x} \d x &= \ln|x| + K,\\ \int \frac{-x}{x^2+1} \d x &= \frac{-\ln|x^2+1|}{2} +K,\\ \int \frac{-x}{(x^2+1)^2} \d x &= \frac{1}{2(x^2+1)}+K. \end{align*}$ So $\begin{align*} \int &\frac{1}{x^5 + 2x^3 + x}\d x \\ &=\ln|x| + \frac{-\ln|x^2+1|}{2} + \frac{1}{2(x^2+1)}+K. \end{align*}$

The general technique for repeated quadratic factors

Suppose you wish to compute $\int \frac {p(x)}{q(x)} \d x$ 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: $q(x) = (ax^2 + b x + c)^n$ then we can always write $\frac {p(x)}{q(x)} = \frac {A_1x+B_1}{ax^2 + bx + c} + \frac {A_2x+B_2}{(ax^2 + bx + c)^2} + \cdots + \frac {A_nx+B_n}{(ax^2 + b x + c)^n}.$ The right-hand side of the equation can be antidifferentiated, though it is not always “easy.”

Reducing rational functions

When computing $\int \frac {p(x)}{q(x)} \d x$ 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.

Compute: $\int \frac {x^3+1}{x^2-1} \d x$

We start by using long-division to reduce the numerator:

Now our integral becomes $\int \frac {x^3+1}{x^2-1}\d x = \int x+ \frac {x+1}{x^2-1}\d x$ From here, we can use our techniques from before to complete the computation, or we can note more simply that

$\frac {x+1}{x^2-1}=\frac {x+1}{(x+1)(x-1)} = \frac {1}{x-1}.$

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.