In this section we investigate the antiderivatives of rational functions. Recall that rational functions are functions of the form $f(x)= \frac{p(x)}{q(x)}\text{,}$ where $p(x)$ and $q(x)$ are polynomials and $q(x)\neq 0\text{.}$ Such functions arise in many contexts, one of which is the solving of certain fundamental differential equations.

We begin with an example that demonstrates the motivation behind this section. Consider the integral $\ds\int \frac{1}{x^2-1}\ dx\text{.}$ We do not have a simple formula for this (if the denominator were $x^2+1\text{,}$ we would recognize the antiderivative as being the arctangent function). It can be solved using Trigonometric Substitution, but note how the integral is easy to evaluate once we realize: \begin{equation*} \frac{1}{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1}. \end{equation*}

Thus \begin{align*} \int\frac{1}{x^2-1}\ dx \amp = \int\frac{1/2}{x-1}\ dx - \int\frac{1/2}{x+1}\ dx\\ \amp = \frac12\ln\abs{x-1} - \frac12\ln\abs{x+1} + C. \end{align*}

This section teaches how to decompose \begin{equation*} \frac{1}{x^2-1} \text{ into } \frac{1/2}{x-1}-\frac{1/2}{x+1}. \end{equation*}

We start with a rational function $f(x)=\frac{p(x)}{q(x)}\text{,}$ where $p$ and $q$ do not have any common factors and the degree of $p$ is less than the degree of $q\text{.}$ It can be shown that any polynomial, and hence $q\text{,}$ can be factored into a product of linear and irreducible quadratic terms. The following Key Idea states how to decompose a rational function into a sum of rational functions whose denominators are all of lower degree than $q\text{.}$

##### Key Idea6.5.1Partial Fraction Decomposition

Let $\ds \frac{p(x)}{q(x)}$ be a rational function, where the degree of $p$ is less than the degree of $q\text{.}$

1. Linear Terms: Let $(x-a)$ divide $q(x)\text{,}$ where $(x-a)^n$ is the highest power of $(x-a)$ that divides $q(x)\text{.}$ Then the decomposition of $\frac{p(x)}{q(x)}$ will contain the sum \begin{equation*} \frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \cdots +\frac{A_n}{(x-a)^n}. \end{equation*}

2. Quadratic Terms: Let $x^2+bx+c$ be an irreducible quadratic that divides $q(x)\text{,}$ where $(x^2+bx+c)^n$ is the highest power of $x^2+bx+c$ that divides $q(x)\text{.}$ Then the decomposition of $\frac{p(x)}{q(x)}$ will contain the sum \begin{equation*} \frac{B_1x+C_1}{x^2+bx+c}+\frac{B_2x+C_2}{(x^2+bx+c)^2}+\cdots+\frac{B_nx+C_n}{(x^2+bx+c)^n}. \end{equation*}

To find the coefficients $A_i\text{,}$ $B_i$ and $C_i\text{:}$

1. Multiply all fractions by $q(x)\text{,}$ clearing the denominators. Collect like terms.

2. Equate the resulting coefficients of the powers of $x$ and solve the resulting system of linear equations.

An irreducible quadratic is a quadratic that has no real solutions. Solving $ax^2+bx+c=0$ using the quadratic equation will determine if a quadratic is irreducible. Completing the square (which is a common integration technique) will also tell you if a quadratic is irreducible.

The following examples will demonstrate how to put this Key Idea into practice. Example 6.5.2 stresses the decomposition aspect of the Key Idea.

##### Example6.5.2Decomposing into partial fractions

Decompose $\ds f(x)=\frac{1}{(x+5)(x-2)^3(x^2+x+2)(x^2+x+7)^2}$ without solving for the resulting coefficients.

Solution
##### Example6.5.3Decomposing into partial fractions

Perform the partial fraction decomposition of $\ds \frac{1}{x^2-1}\text{.}$

Solution

There is another method for finding the partial fraction decomposition called the “Heaviside” method, named after Oliver Heaviside. We show a variation of this process using the same example as in Example 6.5.2.

##### Example6.5.4Decomposing into partial fractions using the Heaviside method

Perform the partial fraction decomposition of $\ds \frac{1}{x^2-1}\text{.}$

Solution
##### Example6.5.5Integrating using partial fractions

Use partial fraction decomposition to integrate $\ds\int\frac{1}{(x-1)(x+2)^2}\ dx\text{.}$

Solution
##### Example6.5.6Integrating using partial fractions

Use partial fraction decomposition to integrate $\ds \int \frac{x^3}{(x-5)(x+3)}\ dx\text{.}$

Solution
##### Example6.5.7Integrating using partial fractions

Use partial fraction decomposition to evaluate $\ds \int\frac{7x^2+31x+54}{(x+1)(x^2+6x+11)}\ dx\text{.}$

Solution

Partial Fraction Decomposition is an important tool when dealing with rational functions. Note that at its heart, it is a technique of algebra, not calculus, as we are rewriting a fraction in a new form. Regardless, it is very useful in the realm of calculus as it lets us evaluate a certain set of “complicated” integrals.

Section 6.6introduces new functions, called the Hyperbolic Functions. They will allow us to make substitutions similar to those found when studying Trigonometric Substitution, allowing us to approach even more integration problems.

# Subsection6.5.1Exercises

Terms and Concepts

In the following exercises, evaluate the indefinite integral.