Setting Up Partial Fractions When You Have Repeated Quadratic Factors - dummies

# Setting Up Partial Fractions When You Have Repeated Quadratic Factors

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 quadratic factors.

This is your worst nightmare when it comes to partial fractions, because the denominator includes repeated quadratic factors.

For each squared quadratic 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 quadratic factor is raised to the nth power, add n partial fractions. For example:

This denominator has one nonrepeating linear factor (x – 8), one nonrepeating quadratic factor (x2 + x + 1), and one quadratic expression that’s squared (x2 + 3). Here’s how you set up the partial fractions:

This example adds one partial fraction for each of the nonrepeating factors and two partial fractions for the squared factor.

When you start out with a quadratic factor of the form (ax2 + C), using partial fractions results in the following two integrals:

Integrate the first by using the variable substitution u = ax2 + C so that du = 2ax dx and

This substitution results in the following integral:

Here are some examples:

To evaluate the second integral, use the following formula:

Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral:

Okay, that’s way too many letters and not nearly enough numbers. Here’s an example:

This is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution u = x2 + 6x + 13 so that du = (2x + 6) dx. If the numerator were 2x + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2:

Because you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator:

This time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two:

At this point, you can use variable substitution to change the first integral as follows:

To solve the second integral, complete the square in the denominator: Divide the b term (6) by 2 and square it, and then represent the C term (13) as the sum of this and whatever’s left:

Now split the denominator into two squares:

To evaluate this integral, use the same formula from the previous section:

So here’s the final answer for the second integral:

Therefore, piece together the complete answer as follows: