Setting Up Partial Fractions When You Have Repeated Linear Factors

By Mark Zegarelli

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated linear factors. These are difficult to work with because each factor requires more than one partial fraction.

For each squared linear factor in the denominator, add two partial fractions in the following form:


For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form:


Generally speaking, when a linear factor is raised to the nth power, add n partial fractions. For example, suppose that you want to integrate the following expression:


This expression contains all linear factors, but one of these factors (x + 5) is nonrepeating and the other (x – 1) is raised to the third power. Set up your partial fractions this way:


Which will yield:


As you can see, this example adds one partial fraction to account for the nonrepeating factor and three to account for the repeating factor.

When you start out with a linear factor, using partial fractions leaves you with an integral in the following form:


Integrate all these cases by using the variable substitution u = ax + b so that du = a dx and


This substitution results in the following integral:


Here are a few examples: