Integrating Using Linear Partial Fractions (BC ONLY) - AP Calculus BC Study Guide

The Core Idea: Integrating Using Linear Partial Fractions (BC ONLY)

The method of partial fractions is an algebraic technique used to integrate rational functions (a ratio of two polynomials). The fundamental problem this topic solves is that while we can easily integrate simple fractions like $\frac{1}{x-2}$, integrating more complex rational functions like $\frac{3x+1}{x^2-x-6}$ is not straightforward.

The core idea is to decompose a complicated rational function into a sum of simpler rational functions that are easily integrated. By breaking the original function down, we can transform a difficult integration problem into a sum of several basic integration problems, which typically result in natural logarithm or simple power rule antiderivatives. This method is essential for expanding the types of functions we can integrate analytically.

The Method of Partial Fractions

The structure of the partial fraction decomposition depends entirely on the factors of the denominator of the rational function. The following rules, derived from the essential knowledge, dictate how to set up the decomposition.

Prerequisite: Polynomial Long Division

Before attempting to decompose a rational function $\frac{P(x)}{Q(x)}$, you must check the degrees of the polynomials.

• Rule: If the degree of the numerator $P(x)$ is greater than or equal to the degree of the denominator $Q(x)$, you must first perform polynomial long division.

• Result: The division will yield a polynomial (or a constant) plus a new rational function where the numerator's degree is strictly less than the denominator's. The method of partial fractions is then applied to this new, proper rational function.

• Form:$$\frac{P(x)}{Q(x)}=(\text{Polynomial})+\frac{R(x)}{Q(x)}$$ where $\mathrm{deg}(R)<\mathrm{deg}(Q)$.

Case 1: Distinct Linear Factors

• Condition: The denominator can be factored into a product of unique linear factors.

• Decomposition Rule: For each distinct linear factor of the form $(ax+b)$ in the denominator, the partial fraction decomposition will include a term $\frac{A}{ax+b}$, where $A$ is a constant to be determined.

• Example Form: If the denominator is $(x-r_1)(x-r_2)...(x-r_n)$, the decomposition is:

$$\frac{P(x)}{(x-r_1)(x-r_2)...(x-r_n)}=\frac{A_1}{x-r_1}+\frac{A_2}{x-r_2}+...+\frac{A_n}{x-r_n}$$

Case 2: Repeated Linear Factors

• Condition: The denominator has a linear factor that is repeated $k$ times, i.e., a factor of the form $(ax+b{)}^k$.

• Decomposition Rule: For each repeated linear factor $(ax+b{)}^k$, the partial fraction decomposition must include a sum of $k$ rational functions. The denominators will be that linear factor raised to the powers of $1,2,\dots ,k$.

• Example Form: If the denominator contains the factor $(x-r{)}^3$, the corresponding part of the decomposition is:

$$\frac{A}{x-r}+\frac{B}{(x-r{)}^2}+\frac{C}{(x-r{)}^3}$$

Understanding the Decomposition Process

The method of partial fractions is a purely algebraic process that must be completed before any integration occurs. The goal is to find the unknown constants ($A,B,C,\dots$) in the decomposition.

The general process for finding these constants is as follows:

1. Set up the Decomposition: Based on the factors of the denominator, write the rational function as a sum of simpler fractions with unknown constants in the numerators, following the rules for distinct and repeated linear factors.

2. Clear the Denominator: Multiply both sides of the equation by the original denominator. This will eliminate all fractions, leaving you with a polynomial equation.

3. Solve for the Constants: There are two common algebraic methods to solve for the constants:

   • Method A: Equating Coefficients: Expand the polynomial on the side with the unknown constants. Group terms by powers of $x$ (e.g., all $x^2$ terms, all $x$ terms, and all constant terms). Since the equation must hold for all values of $x$, the coefficients of corresponding powers of $x$ on both sides must be equal. This creates a system of linear equations that you can solve for the constants.

   • Method B: Substituting Convenient Values: Since the polynomial equation from Step 2 must be true for all values of $x$, you can substitute convenient values of $x$ to simplify the equation and solve for the constants. The most convenient values are the roots of the original denominator, as they will cause many terms to become zero.

Once the constants are found, substitute them back into the decomposition. The original integral is now expressed as the integral of the sum of these simpler terms.

Core Concepts & Rules

• Purpose: The method of partial fractions is a technique for rewriting a complex rational function into a sum of simpler, more easily integrated rational functions.

• Prerequisite Check: Always check the degrees of the numerator and denominator first. If the degree of the numerator is greater than or equal to the degree of the denominator, you must perform polynomial long division before proceeding.

• Distinct Linear Factors: For every distinct linear factor $(ax+b)$ in the denominator, the decomposition must include a term of the form $\frac{A}{ax+b}$.

• Repeated Linear Factors: For every factor $(ax+b{)}^k$ in the denominator, the decomposition must include $k$ separate terms: $\frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+\cdots +\frac{A_k}{(ax+b{)}^k}$.

• Integration: After decomposition, the resulting terms are integrated. Terms of the form $\frac{A}{ax+b}$ integrate to logarithmic functions, while terms of the form $\frac{A}{(ax+b{)}^n}$ for $n>1$ are integrated using the power rule.

Step-by-Step Example 1: Distinct Linear Factors

Problem: Find $\int \frac{x+7}{x^2-x-6}dx$.

Step 1: Factor the Denominator

The denominator is $x^2-x-6$, which factors into $(x-3)(x+2)$. These are distinct linear factors. The degree of the numerator (1) is less than the degree of the denominator (2), so long division is not needed.

Step 2: Set Up the Partial Fraction Decomposition

Based on the distinct linear factors, the decomposition will have the form:

$$\frac{x+7}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$$

Step 3: Solve for the Constants $A$ and $B$

Multiply both sides by the original denominator, $(x-3)(x+2)$, to clear the fractions:

$$x+7=A(x+2)+B(x-3)$$

Now, we can solve for $A$ and $B$ by substituting convenient values for $x$.

• Let $x=3$. This will make the $B$ term zero.

  $$3+7=A(3+2)+B(3-3)$$

  $$10=5A⟹A=2$$

• Let $x=-2$. This will make the $A$ term zero.

  $$-2+7=A(-2+2)+B(-2-3)$$

  $$5=-5B⟹B=-1$$

Step 4: Rewrite and Solve the Integral

Substitute the values of $A$ and $B$ back into the decomposition:

$$\int \frac{x+7}{x^2-x-6}dx=\int \left(\frac{2}{x-3}-\frac{1}{x+2}\right)dx$$

Now, integrate each term separately:

$$=2\int \frac{1}{x-3}dx-\int \frac{1}{x+2}dx$$

$$=2\ln |x-3|-\ln |x+2|+C$$

Step-by-Step Example 2: Repeated Linear Factors and Long Division

Problem: Find $\int \frac{x^3-x^2-5x+7}{x^2-2x+1}dx$.

Step 1: Perform Long Division

The degree of the numerator (3) is greater than the degree of the denominator (2). We must perform long division.

Note that the denominator $x^2-2x+1=(x-1{)}^2$.

x + 1 ________________ x^2-2x+1 | x^3 - x^2 - 5x + 7 -(x^3 - 2x^2 + x) _________________ x^2 - 6x + 7 -(x^2 - 2x + 1) _____________ -4x + 6

The result of the division is $x+1$ with a remainder of $-4x+6$. So, the original integral can be rewritten as:

$$\int \left(x+1+\frac{-4x+6}{x^2-2x+1}\right)dx=\int (x+1)dx+\int \frac{-4x+6}{(x-1{)}^2}dx$$

Step 2: Decompose the Remainder Fraction

Now we apply partial fractions to $\frac{-4x+6}{(x-1{)}^2}$. The denominator is a repeated linear factor $(x-1{)}^2$.

The decomposition is:

$$\frac{-4x+6}{(x-1{)}^2}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}$$

Step 3: Solve for the Constants $A$ and $B$

Multiply by $(x-1{)}^2$ to clear the denominator:

$$-4x+6=A(x-1)+B$$

• Let $x=1$. This will make the $A$ term zero.

  $$-4(1)+6=A(1-1)+B$$

  $$2=B$$

• Now substitute $B=2$ back into the equation:

  $$-4x+6=A(x-1)+2$$

  To find $A$, we can either expand and equate coefficients or substitute any other value for $x$. Let's use $x=0$.

  $$-4(0)+6=A(0-1)+2$$

  $$6=-A+2⟹4=-A⟹A=-4$$

Step 4: Rewrite and Solve the Full Integral

Substitute the constants back into the integral expression from Step 1:

$$\int (x+1)dx+\int \left(\frac{-4}{x-1}+\frac{2}{(x-1{)}^2}\right)dx$$

Integrate each part:

$$=\left(\frac{x^2}{2}+x\right)-4\int \frac{1}{x-1}dx+2\int (x-1{)}^{-2}dx$$

$$=\frac{x^2}{2}+x-4\ln |x-1|+2\frac{(x-1{)}^{-1}}{-1}+C$$

$$=\frac{x^2}{2}+x-4\ln |x-1|-\frac{2}{x-1}+C$`

Using Your Calculator

The method of partial fractions is a purely analytical technique for finding an antiderivative. A graphing calculator cannot perform the symbolic algebra required for the decomposition.

The calculator's primary role for this topic is to verify your work on a definite integral.

1. Solve Analytically: First, find the antiderivative by hand using the steps outlined above.

2. Evaluate: If the problem is a definite integral, use the Fundamental Theorem of Calculus to evaluate your antiderivative at the limits of integration to find the exact numerical answer.

3. Verify with Calculator: Use your calculator's numerical integration feature (e.g., fnInt on a TI-84 or the equivalent menu option) to calculate the value of the definite integral of the original function.

4. Compare: The decimal approximation from your calculator should match the value you calculated by hand. If they do not match, there is likely an error in your algebraic decomposition or your integration.

For example, to check ${\int }_4^5\frac{x+7}{x^2-x-6}dx$, you would first find the antiderivative $2\ln |x-3|-\ln |x+2|$. Then evaluate $[2\ln |2|-\ln |7|]-[2\ln |1|-\ln |6|]=2\ln (2)-\ln (7)+\ln (6)$. Finally, you would use fnInt((X+7)/(X^2-X-6), X, 4, 5) on your calculator and compare the decimal result to your exact answer.

AP Exam Quick Hit

Common Question Types

• Direct Antiderivative Calculation: A multiple-choice or free-response question may directly ask for the integral of a rational function that requires partial fractions.

  • Example: "Find $\int \frac{5}{x^2+x-6}dx$."

• Solving a Separable Differential Equation: This is a very common application in free-response questions. After separating variables, the integration step for one of the variables requires partial fractions.

  • Example: "Find the particular solution to $\frac{dy}{dx}=\frac{y}{x^2+3x+2}$ with the initial condition $y(0)=3$." You would separate to get $\frac{1}{y}dy=\frac{1}{x^2+3x+2}dx$, and the integral on the right requires partial fractions.

Common Mistakes

• Forgetting Long Division: Students often immediately try to decompose a fraction like $\frac{x^3}{x^2-1}$ without first performing long division. This is incorrect because the method only works when the degree of the numerator is strictly less than the degree of the denominator.

• Incorrect Setup for Repeated Roots: A very common error is to set up the decomposition for a denominator like $(x-5{)}^2$ as $\frac{A}{(x-5{)}^2}$ only. The correct setup must include a term for each power: $\frac{A}{x-5}+\frac{B}{(x-5{)}^2}$.

• Algebraic Errors: Simple mistakes made when solving for the constants $A,B,C,\dots$ are frequent. This includes errors in distributing, combining like terms, or solving the resulting system of equations.

• Integration Errors: After a correct decomposition, students might incorrectly integrate the resulting terms. A common mistake is treating $\int \frac{1}{(x-c{)}^2}dx$ as a logarithm; it should be integrated using the power rule to get $-(x-c{)}^{-1}+C$. Another is forgetting the absolute value inside the natural logarithm, $\ln |u|$.