3 min readβ’june 18, 2024

Catherine Liu

Jed Quiaoit

AP Description & Expectations

**Enduring Understanding FUN-6:**Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.

**Essential Knowledge FUN-6.F.1:**Some rational functions can be decomposed into sums of ratios of linear, non-repeating factors to which basic integration techniques can be applied. π‘

Integrating** rational functions **can be a challenging task, but one powerful technique for solving these types of integrals is the method of partial fractions. This method allows us to **decompose** a rational function into a sum of simpler functions that can be integrated individually. π§

To begin, we first need to factor the denominator of the rational function into its distinct linear factors. If the denominator is not factorable, we can use long division to convert the rational function into a polynomial plus a remainder, in which the remainder is divided by a factorable denominator.

Next, we use the method of undetermined coefficients to find the coefficients of the partial fraction decomposition. This involves setting up a system of equations by multiplying both sides of the decomposition by the denominator and equating it to the original function. Then we solve this system of equations to find the coefficients.

For example, consider the rational function f(x) = (x^2 + 2x + 1)/(x^3 + x^2 - 6x - 6). We can factor the denominator as (x^3 + x^2 - 6x - 6) = (x+3)(x+2)(x-1). We can then use the method of undetermined coefficients to find the coefficients of the partial fraction decomposition:

Once we have the coefficients, we can integrate each of the individual terms in the decomposition. For example, integrating A(x+3) would give us A(x+3)^2/2. We can then add these individual integrals to find the final solution.

It's important to note that some cases may require the use of long division or repeated partial fraction. This happens when the degree of the numerator is equal or greater than the degree of the denominator, in these cases, we divide the numerator by the denominator and find the remainder, then we divide the remainder by the denominator and find a new remainder, and so on. Each remainder will have a partial fraction, and by summing all these fractions we will get the final solution. π€

Let's consider the following integral:

The integrand is a rational function with two linear factors in the denominator. With our current techniques, we canβt solve this integral. However, by using partial fractions, we can split the rational integrand into two separate fractions that we know how to handle.

For the AP Calculus BC exam, you will only be dealing with examples where the denominator can be factored into non-repeating linear factors. A non-repeating linear factor is a term of the form (Ax+B) where A and B are numbers, and each term only shows up once in the denominator. π

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πUnit 1 β Limits & Continuity

π€Unit 2 β Fundamentals of Differentiation

π€π½Unit 3 β Composite, Implicit, & Inverse Functions

πUnit 4 β Contextual Applications of Differentiation

β¨Unit 5 β Analytical Applications of Differentiation

π₯Unit 6 β Integration & Accumulation of Change

πUnit 7 β Differential Equations

πΆUnit 8 β Applications of Integration

π¦Unit 9 β Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)

βΎUnit 10 β Infinite Sequences & Series (BC Only)

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