The Method of Partial Fractions

Section 7.5 The Method of Partial Fractions

In this section, we will learn an algebraic method called the partial fraction decomposition to help evaluate integrals.

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Objectives

After this section, students will be able to:

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determine whether a rational expression is proper or improper
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set up the correct form of the partial fraction decomposition based on the factorization of the denominator
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find the constants in a partial fraction decomposition
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evaluate integrals of rational expressions using the method of partial fraction decomposition.
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Investigation 7.5.1.

Evaluate the indefinite integral \(\displaystyle \int\dfrac{5x - 4}{2x^2 + x - 1}\, dx\text{.}\) Can we evaluate it using any of the methods we have learned so far in the class?

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If you get stuck and/or ready to be helped. Click open Richard’s hint below.

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Hint.

What is \(\dfrac{3}{x + 1} - \dfrac{1}{2x - 1}\text{?}\)

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How do we subtract fractions?

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Solution.

One can verify quickly that

\begin{align*} \frac{3}{x + 1} - \frac{1}{2x - 1} \amp= \frac{3\cdot(2x - 1)}{(x + 1)\cdot(2x - 1)} - \frac{1\cdot(x + 1)}{(2x - 1)\cdot(x + 1)} \\ \amp= \frac{3(2x - 1) - (x + 1)}{(x + 1)(2x - 1)} \\ \amp= \frac{6x - 3 - x - 1}{2x^2 - x + 2x - 1} \\ \amp= \frac{5x - 4}{2x^2 + x - 1} \end{align*}

This implies that \(\dfrac{5x - 4}{2x^2 + x - 1} = \dfrac{3}{x + 1} - \dfrac{1}{2x - 1}\text{.}\)

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But why is it helpful... That is because we can now split up the integral (using linearity) and integrate each of the smaller fractions.

\begin{align*} \int\dfrac{5x - 4}{2x^2 + x - 1}\, dx \amp= \int\left(\dfrac{3}{x + 1} - \dfrac{1}{2x - 1}\right)\, dx\\ \amp= 3\int \frac{1}{x + 1}\, dx - \int \frac{1}{2x - 1}\, dx \amp\amp\text{by linearity} \\ \amp= 3\ln\left|x + 1\right| - \frac{1}{2}\ln\left|2x - 1\right| + C \amp\amp\text{by quick little -sub} \end{align*}

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Observe that the key step here that allows us to evaluate the integral is captured in Richard’s hint that he somehow knew how to decompose a giant fraction in the integrand into two smaller terms. This section will introduce this fancy method of decomposing a giant fraction into smaller terms, called the partial fraction decomposition.

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This is essentially an algebraic method, in the sense that we don’t need any calculus to make sense of this method. After we decompose the giant fraction into smaller terms, we still need to integrate each smaller term of the fractions.

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Subsection Prerequisite: Proper Rational Expression

In order to apply the method of partial fraction decomposition, we need the fraction to be proper. But what does a "proper fraction" mean...

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In short, a proper fraction has a smaller numerator than the denominator.

\begin{align*} \text{Examples of Proper Fractions} \amp: \frac{3}{8}, \, \frac{1}{4}, \, \frac{14}{15}, \, \frac{4}{5} \\ \amp \\ \text{Examples of Improper Fractions} \amp: \frac{4}{3}, \, \frac{11}{4}, \, \frac{7}{7} \end{align*}

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The reason why we call such fractions the proper fractions is because it makes sense to visualize the idea of "part of a whole" in these types of fractions, where the denominator of a fraction represents the total number of parts in a whole and the numerator represents the parts we are taking. In other words, the value of a proper fraction must be between \(0\) and \(1\text{.}\)

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Similarly, a proper rational expression is defined as a rational expression where the numerator is less than the denominator. But how do we compare polynomials though (for example, which one is greater, \(x^3 - 1\) or \(2x^2 + 5\text{?}\)). We usually compare the degree of the polynomials (the highest power on \(x\)).

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That is, proper rational expressions are the ones where the degree of the numerator is less than the degree of the denominator.

\begin{align*} \text{Examples of Proper Rational Expression} \amp: \frac{x^2 - 3x + 7}{x^4 - 16}, \, \frac{x^2 - 5}{x^3 + 2x}, \, \frac{2x + 9}{x^4}, \, \frac{x - 1}{x^2 - 9} \\ \amp \\ \text{Examples of Improper Rational Expression} \amp: \frac{2x^2 + 7}{x - 5}, \, \frac{x - 2}{x - 5}, \, \frac{x^3}{6x^2 + 2x - 1} \end{align*}

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The method of partial fraction decomposition will only work on proper rational expressions.

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Subsection Partial Fraction Decomposition: Distinct Linear Factors

Recall the idea of the partial fraction decomposition is to decompose a fraction into smaller terms. If we know what the denominator can be factored into, then we know the denominators of the little fractions.

\begin{equation*} \frac{p(x)}{(x - r_1)(x - r_2)} = \frac{?}{x - r_1} + \frac{?}{x - r_2} \end{equation*}

So the work here is really to determine the mysterious numerators. There are multiple strategies to determine the numerators. We will first look at an example with specific rational expressions to demonstrate the two main strategies to determine the numerators.

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Example 7.5.1.

Evaluate the indefinite integral \(\displaystyle \int \frac{3x}{x^2 + 2x - 8}\, dx\)

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Hint.

Observe that the integrand is a proper rational expression, which means we may be able to use the method of partial fraction decomposition. This will allow us to decompose the integrand into smaller terms of fractions that we can integrate easily.

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To determine the denominators of the smaller terms of fractions, we need to factor this denominator.

\begin{equation*} x^2 + 2x - 8 = (x + 4)(x - 2) \end{equation*}

This implies that

\begin{equation*} \frac{3x}{x^2 + 2x - 8} = \frac{?}{x + 4} + \frac{?}{x - 2} \end{equation*}

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We don’t know the numerators yet (that is why Richard put ?’s on the fractions). Let’s call them \(A\) and \(B\text{,}\) where they are two variables. Then we are essentially solving the following rational equation for \(A\) and \(B\text{:}\)

\begin{equation*} \frac{3x}{x^2 + 2x - 8} = \frac{A}{x + 4} + \frac{B}{x - 2} \end{equation*}

Now put your MTH 95 thinking cap on. How do we solve a rational equation?

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Solution.

Let’s first apply the method of partial fraction decomposition to the integrand first to decompose the proper rational expression.

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Let \(A\) and \(B\) be two constants such that

\begin{equation*} \frac{3x}{(x + 4)(x - 2)} = \frac{A}{x + 4} + \frac{B}{x - 2} \end{equation*}

Then the goal here is to solve this equation for \(A\) and \(B\text{.}\) One strategy to solve a rational equation is to multiply out the denominators.

\begin{align*} (x + 4)(x - 2)\cdot \frac{3x}{(x + 4)(x - 2)} \amp= \left(\frac{A}{x + 4} + \frac{B}{x - 2}\right)\cdot (x + 4)(x - 2) \\ \amp \\ 3x \amp= A(x - 2) + B(x + 4) \end{align*}

To determine the values of \(A\) and \(B\text{,}\) there are two methods here.

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Method I: Determining Coefficients. We can simplify the right-hand side of the equation and combine the like terms.

\begin{equation*} 3x = (A + B)x + (-2A + 4B) \end{equation*}

Two functions are the same when they look exactly the same, including the coefficients. This equation implies that

\begin{equation*} \begin{cases} A + B = 3 \amp \\ -2A + 2B = 6 \amp \end{cases} \qquad\implies \qquad \begin{cases} A \amp= 2 \\ B \amp= 1 \end{cases} \end{equation*}

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Method II: Special Values (or Cover-up Method). We know the equation

\begin{equation*} 3x = A(x - 2) + B(x + 4) \end{equation*}

is true for all \(x\)’s. So we want to plug in some values of \(x\)’s to "cover up" either \(A\) or \(B\text{.}\)

To determine the value of \(A\text{,}\) we need to cover up \(B\text{,}\) which means we want the coefficient of \(B\) to be \(0\text{.}\) That is, we will let \(x = -4\text{.}\) Then

\begin{align*} 3(-4) \amp= A(-4 - 2) + B(-4 + 4) \\ -12 \amp= -6A \\ A \amp= 2 \end{align*}

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To determine the value of \(B\text{,}\) we need to cover up \(A\text{,}\) which means we want the coefficient of \(A\) to be \(0\text{.}\) That is, we will let \(x = 2\text{.}\) Then

\begin{align*} 3(2) \amp= A(2 - 2) + B(2 + 4) \\ 6 \amp= 6B \\ B \amp= 1 \end{align*}

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Hence, we know that \(A = 2\) and \(B = 1\text{.}\)

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It doesn’t matter which method you used to determine \(A\) and \(B\text{,}\) we know that \(A = 2\) and \(B = 1\text{.}\) This implies that

\begin{equation*} \frac{3x}{(x + 4)(x - 2)} = \frac{2}{x + 4} + \frac{1}{x - 2} \end{equation*}

Then we obtain

\begin{align*} \int \frac{3x}{(x + 4)(x - 2)}\, dx \amp= \int \left(\frac{2}{x + 4} + \frac{1}{x - 2}\right)\, dx \\ \amp= 2\int \frac{1}{x + 4}\, dx + \int \frac{1}{x - 2} \, dx \amp\amp\text{by linearity} \\ \amp= 2\ln\left|x + 4\right| + \ln\left|x - 2\right| + C \end{align*}

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Observe that we will need to determine the form of the partial fraction decomposition in the setup. Then we solve this rational equation by multiplying the giant denominator on both sides of the equation. Next, we will determine the values of the variables using Method I: Determining Coefficients or Method II: Special Values (or Cover-up Method). This will allow us to decompose the giant fraction into smaller terms, and we can integrate each smaller term by linearity.

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We will summarize the form of the partial fraction decomposition into a nice little list (there are three cases in total).

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To determine the value of the variables, it is up to you which method, I or II, you want to use. They have their advantages and shortcomings so Richard will not make you use one or the other (but you will need to understand at least one of the two methods).

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Example 7.5.2.

Evaluate the indefinite integral \(\displaystyle \int \frac{6x + 4}{x^2 - 1}\, dx\text{.}\)

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Hint.

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So determine the form of the partial fraction decomposition in the setup using \(A\) and \(B\) as the variables. Then solve the equation for \(A\) and \(B\) to decompose the integrand. Next, integrate the smaller terms of fractions.

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Solution.

Observe we can factor the denominator

\begin{equation*} x^2 - 1 = (x + 1)(x - 1) \end{equation*}

Then the form of the partial fraction decomposition in the setup is

\begin{equation*} \frac{6x + 4}{(x + 1)(x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1} \end{equation*}

where \(A\) and \(B\) are constants.

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We can solve this rational equation by multiplying the giant denominator on both sides of the equation. We obtain

\begin{align*} (x + 1)(x - 1)\cdot \frac{6x + 4}{(x + 1)(x - 1)} \amp= \left(\frac{A}{x + 1} + \frac{B}{x - 1}\right)\cdot (x + 1)(x - 1) \\ \amp \\ 6x + 4 \amp= A(x - 1) + B(x + 1) \end{align*}

We can determine the values of \(A\) and \(B\) using either method I or method II.

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Method I: Determining Coefficients. We can simplify the right-hand side of the equation and combine like terms.

\begin{equation*} 6x + 4 = (A + B)x + (-A + B) \end{equation*}

Now we are matching up the coefficients. This implies that

\begin{equation*} \begin{cases} A + B = 6 \amp \\ -A + B = 4 \amp \end{cases} \qquad\implies \qquad \begin{cases} A \amp= 1 \\ B \amp= B \end{cases} \end{equation*}

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Method II: Special Values (or Cover-up Method). We want to plug in some values of \(x\)’s to "cover up" either \(A\) or \(B\text{.}\)

To determine the value of \(A\text{,}\) we need to cover up \(B\text{.}\) To do so, we let \(x = -1\text{.}\) Then

\begin{align*} 6(-1) + 4 \amp= A(-1 - 1) + B(-1 + 1) \\ -2 \amp= -2A \\ A \amp= 1 \end{align*}

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To determine the value of \(B\text{,}\) we need to cover up \(A\text{.}\) To do so, we let \(x = 1\text{.}\) Then

\begin{align*} 6(1) + 4 \amp= A(1 - 1) + B(1 + 1) \\ 10 \amp= 2B \\ B \amp= 5 \end{align*}

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Now we know the values of \(A\) and \(B\text{,}\) we obtain

\begin{equation*} \frac{6x + 4}{(x + 1)(x - 1)} = \frac{1}{x + 1} + \frac{5}{x - 1} \end{equation*}

This implies that

\begin{align*} \int \frac{6x + 4}{(x + 1)(x - 1)}\, dx \amp= \int \left(\frac{1}{x + 1} + \frac{5}{x - 1} \right)\, dx \\ \amp= \int \frac{1}{x + 1}\, dx + \int 5\frac{1}{x - 1}\, dx \amp\amp\text{by linearity}\\ \amp= \ln\left|x + 1\right| + 5\ln\left|x - 1\right| + C \end{align*}

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But what if the denominator can be factored into more than two distinct factors? The setup is similar. We will just need as many fractions as the number of distinct factors in the setup, with each fraction taking one factor as the denominator.

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Example 7.5.3.

Evaluate the indefinite integral \(\displaystyle \int \frac{3x^2 + 7x - 2}{x^3 - x^2 - 2x}\, dx\)

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Hint.

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Furthermore, we can factor the denominator as follows

\begin{equation*} x^3 - x^2 - 2x = x(x - 2)(x + 1) \end{equation*}

Then we can decompose the giant fraction in the integral into three little fractions where each fraction has the denominator of a factor.

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Solution.

Based on the factors, the form of the partial fraction decomposition in the setup is

\begin{equation*} \frac{3x^2 + 7x - 2}{x(x - 2)(x + 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1} \end{equation*}

where \(A\text{,}\) \(B\text{,}\) and \(C\) are constants.

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We can solve this rational equation by multiplying the giant denominator on both sides of the equation. We obtain

\begin{align*} x(x - 2)(x + 1)\cdot\frac{3x^2 + 7x - 2}{x(x - 2)(x + 1)} \amp= \left(\frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}\right)\cdot x(x - 2)(x + 1) \\ \amp \\ 3x^2 + 7x - 2 \amp= A(x - 2)(x + 1) + Bx(x + 1) + Cx(x - 2) \end{align*}

We can determine the values of \(A\text{,}\) \(B\text{,}\) and \(C\) using either method I or method II.

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Richard will present method II in this solution. You can for sure approach this using method I if you wish.

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To determine the value of \(A\text{,}\) we need to cover up both \(B\) and \(C\text{.}\) To do so, we let \(x = 0\text{.}\)

\begin{align*} 3(0)^2 + 7(0) - 2 \amp= A(0 - 2)(0 + 1) + B(0)(0 + 1) + C(0)(0 - 2) \\ -2 \amp= -2A \\ A \amp= 1 \end{align*}

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To determine the value of \(B\text{,}\) we need to cover up both \(A\) and \(C\text{.}\) To do so, we let \(x = 2\text{.}\)

\begin{align*} 3(2)^2 + 7(2) - 2 \amp= A(2 - 2)(2 + 1) + B(2)(2 + 1) + C(2)(2 - 2) \\ 24 \amp= 6B \\ B \amp= 4 \end{align*}

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To determine the value of \(C\text{,}\) we need to cover up both \(A\) and \(B\text{.}\) To do so, we let \(x = -1\text{.}\)

\begin{align*} 3(-1)^2 + 7(-1) - 2 \amp= A(-1 - 2)(-1 + 1) + B(-1)(-1 + 1) + C(-1)(-1 - 2) \\ -6 \amp= 3C \\ C \amp= -2 \end{align*}

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Now we know the values of \(A\text{,}\) \(B\text{,}\) and \(C\text{,}\) we obtain

\begin{equation*} \frac{3x^2 + 7x - 2}{x(x - 2)(x + 1)} = \frac{1}{x} + \frac{4}{x - 2} + \frac{-2}{x + 1} \end{equation*}

This implies that

\begin{align*} \int \frac{3x^2 + 7x - 2}{x(x - 2)(x + 1)}\, dx \amp= \int\left( \frac{1}{x} + \frac{4}{x - 2} + \frac{-2}{x + 1} \right)\, dx \\ \amp= \int \frac{1}{x}\, dx + 4\int \frac{1}{x - 2}\, dx -2\int \frac{1}{x + 1}\, dx \\ \amp= \ln|x| + 4\ln|x - 2| - 2\ln|x + 1| + C \end{align*}

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Subsection Partial Fraction Decomposition: Repeated Factors

A follow-up question to ask here is what if there are repeated factors? Can we use the same setup as the previous example?

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Let’s look at a quick example (we will integrate this fraction later). Let’s say we want to decompose the fraction

\begin{equation*} \frac{5x^2 - 3x + 2}{x^3 - 2x^2} \end{equation*}

This is a proper fraction so we may be able to apply the method of partial fraction decomposition. The denominator can be factored into \(x\cdot x\cdot(x - 2)\text{,}\) so can we set up the form of partial fraction decomposition as follows?

\begin{equation*} \frac{5x^2 - 3x + 2}{x^3 - 2x^2} = \frac{5x^2 - 3x + 2}{x\cdot x\cdot (x - 2)} = \frac{A}{x} + \frac{B}{x} + \frac{C}{x - 2} \end{equation*}

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The short answer is NO because we won’t be able to obtain the denominator of \(x^3 - 2x^2\) when adding the three little terms of the fractions together. Imagine you wanted to add the three little terms of the fractions together, we would obtain

\begin{equation*} \frac{A}{x} + \frac{B}{x} + \frac{C}{x - 2} = \frac{(A + B)(x - 2) + Cx}{x(x - 2)} \end{equation*}

Observe this denominator doesn’t match up with the denominator of our giant fraction.

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One way to fix this issue is to replace one of the denominator of \(x\) with an \(x^2\) to force out a factor of \(x^2\) in the common denominator when adding them together. Then the correct set up for the form of the partial fraction decomposition is

\begin{equation*} \frac{5x^2 - 3x + 2}{x\cdot x\cdot (x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2} \end{equation*}

Now let’s put this in use and integrate the giant fraction!

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Example 7.5.4.

Evaluate the indefinite integral \(\displaystyle \int \frac{5x^2 - 3x + 2}{x^3 - 2x^2}\, dx\)

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Hint.

Now you know the form of the partial fraction decomposition in the setup. Determine the values of \(A\text{,}\) \(B\text{,}\) and \(C\) to help evaluate the integral!

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Solution.

The form of the partial fraction decomposition in the setup is

\begin{equation*} \frac{5x^2 - 3x + 2}{x^2(x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2} \end{equation*}

We can solve this rational equation by multiplying the giant denominator on both sides of the equation. We obtain

\begin{align*} x^2(x - 2)\cdot \frac{5x^2 - 3x + 2}{x\cdot x\cdot (x - 2)} \amp= \left( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2} \right)\cdot x^2(x - 2)\\ \amp \\ 5x^2 - 3x + 2 \amp= Ax(x - 2) + B(x - 2) + Cx^2 \end{align*}

We can determine the values of \(A\text{,}\) \(B\text{,}\) and \(C\) using either method I or method II.

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Richard will present method II in this solution. If you know a little bit of matrix algebra, method I will be a lot easier by the way.

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We can determine the value of \(B\) and \(C\) easily.

To determine the value of \(B\text{,}\) we need to cover up both \(A\) and \(C\text{.}\) To do so, we let \(x = 0\text{.}\)

\begin{align*} 5(0)^2 - 3(0) + 2 \amp= A(0)(0 - 2) + B(0 - 2) + C(0)^2 \\ 2 \amp= -2B \\ B \amp= -1 \end{align*}

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To determine the value of \(C\text{,}\) we need to cover up both \(A\) and \(B\text{.}\) To do so, we let \(x = 2\text{.}\)

\begin{align*} 5(2)^2 - 3(2) + 2 \amp= A(2)(2 - 2) + B(2 - 2) + C(2)^2 \\ 16 \amp= 4C \\ C \amp= 4 \end{align*}

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But there is no way we can cover \(B\) and \(C\) at the same time...

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One way to go around this issue is to plug back the value of \(B\) and \(C\text{,}\) and let \(x\) be any other numbers than \(0\) and \(2\text{.}\) This will result in an equation with JUST \(A\) as the variable. For simplicity, Richard will let \(x = 1\text{.}\) Plugging in \(B = -1\) and \(C = 4\) back to the equation, we obtain

\begin{align*} 5(1)^2 - 3(1) + 2 \amp= A(1)(1 - 2) - 1(1 - 2) + 4(1)^2 \\ 4 \amp= -A + 1 + 4 \\ A \amp= 1 \end{align*}

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Now we know the values of \(A\text{,}\) \(B\text{,}\) and \(C\text{,}\) we obtain

\begin{equation*} \frac{5x^2 - 3x + 2}{x^2(x - 2)} = \frac{1}{x} + \frac{-1}{x^2} + \frac{4}{x - 2} \end{equation*}

This implies that

\begin{align*} \int \frac{5x^2 - 3x + 2}{x^2(x - 2)}\, dx \amp= \int \left( \frac{1}{x} + \frac{-1}{x^2} + \frac{4}{x - 2} \right)\, dx \\ \amp= \int \frac{1}{x}\, dx - \int \frac{1}{x^2}\, dx + 4\int \frac{1}{x - 2}\, dx \\ \amp= \ln|x| + x^{-1} + 4\ln|x - 2| + C \end{align*}

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But Richard... why can’t we factor the denominator of \(x^3 - 2x^2\) into \(x^2(x - 2)\) and set up the form of the partial fraction decomposition as follows

\begin{equation*} \frac{5x^2 - 3x + 2}{x^2(x - 2)} = \frac{A}{x^2} + \frac{B}{x - 2} \end{equation*}

The issue lies on the fraction of \(\dfrac{A}{x^2}\) in the setup.

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Let’s grab the correct decomposition we had earlier and combine the first two terms. Then we obtain

\begin{align*} \frac{1}{x} - \frac{1}{x^2} + \frac{4}{x - 2} \amp= \frac{1\cdot x}{x\cdot x} - \frac{1}{x^2} + \frac{4}{x - 2} \\ \amp= \frac{x}{x^2} - \frac{1}{x^2} + \frac{4}{x - 2} \\ \amp= \frac{x - 1}{x^2} + \frac{4}{x - 2} \end{align*}

Observe that the first fraction doesn’t have a constant on the numerator, but rather a linear factor. Hence, there is no constant \(A\) that will make \(\frac{A}{x^2}\) equal to \(\frac{x - 1}{x^2}\text{.}\)

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Subsection Partial Fraction Decomposition: Irreducible Quadratic Factors

We also know that there are some quadratic expressions we can’t factor. Then how do we set up the form of the partial fraction decomposition with an irreducible factor in the denominator?

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Let’s look at another quick example (we will integrate this fraction later). Let’s say we want to decompose the fraction

\begin{equation*} \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} \end{equation*}

This is also a proper fraction so we may be able to apply the method of partial fraction decomposition. Observe that the denominator is factored completely, which means we can’t factor \(x^2 - 2x + 3\) further (Why is that? This is left as an exercise for the readers to figure out. If you need a hint, the factor theorem will help you out). So can we set up the form of partial fraction decomposition as follows?

\begin{equation*} \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} = \frac{A}{x - 2} + \frac{B}{x^2 - 2x + 3} \end{equation*}

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The short answer is NO because there is no guarantee that the numerator of the second fraction (the one with the denominator of the quadratic expression) is a constant.

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Observe the previous example that we can decompose the fraction of \(\frac{5x^2 - 3x + 2}{x^2(x - 2)}\) as \(\frac{x - 1}{x^2} + \frac{4}{x - 2}\text{.}\) This implies that a quadratic denominator may have a non-constant linear numerator. Since we don’t know whether the numerator of the quadratic denominator is a constant or non-constant linear expression, we can set the numerator to be a generic linear expression of \(Bx + C\) to take into account both possibilities (if the numerator is a constant, then \(B\) would be \(0\text{;}\) if the numerator is not a constant, then \(B\) is not \(0\)). Hence, the correct setup for the form of the partial fraction decomposition is

\begin{equation*} \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 - 2x + 3} \end{equation*}

Now let’s put this in use and integrate the giant fraction!

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Example 7.5.5.

Evaluate the indefinite integral \(\displaystyle \int \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)}\, dx \)

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Hint.

Now you know the form of the partial fraction decomposition in the setup. Determine the values of \(A\text{,}\) \(B\text{,}\) and \(C\) to help evaluate the integral!

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Solution.

The form of the partial fraction decomposition in the setup is

\begin{equation*} \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 - 2x + 3} \end{equation*}

We can solve this rational equation by multiplying the giant denominator on both sides of the equation. We obtain

\begin{align*} (x - 2)\left(x^2 - 2x + 3\right)\cdot \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} \amp= \left(\frac{A}{x - 2} + \frac{Bx + C}{x^2 - 2x + 3}\right)\cdot (x - 2)\left(x^2 - 2x + 3\right) \\ \amp \\ 7x^2 - 16x + 19 \amp= A\left(x^2 - 2x + 3\right) + (Bx + C)(x - 2) \end{align*}

We can determine the values of \(A\text{,}\) \(B\text{,}\) and \(C\) using either method I or method II.

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Richard will present method II in this solution. If you know a little bit of matrix algebra, method I will be a lot easier by the way.

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We can determine the value of \(A\) easily. To determine the value of \(A\text{,}\) we need to cover up both \(B\) and \(C\text{.}\) To do so, we let \(x = 2\text{.}\)

\begin{align*} 7(2)^2 - 16(2) + 19 \amp= A\left(2^2 - 2(2) + 3\right) + (B(2) + C)(2 - 2) \\ 15 \amp= 3A \\ A \amp= 5 \end{align*}

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We can pull the same trick to determine the value of \(B\) and \(C\) by picking another value for \(x\) and plug back \(A = 5\) to the equation. Richard will let \(x = 0\) and plug in \(A = 5\) to the equation. Then we obtain

\begin{align*} 7(0)^2 - 16(0) + 19 \amp= 5\left(0^2 - 2(0) + 3\right) + (B(0) + C)(0 - 2) \\ 19 \amp= 15 - 2A \\ C \amp= 2 \end{align*}

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To determine the value of \(B\text{,}\) Richard will let \(x = 1\) (for simplicity) and plug in \(A = 5\) and \(C = 1\) back to the equation. Then we obtain

\begin{align*} 7(1)^2 - 16(1) + 19 \amp= 5\left(1^2 - 2(1) + 3\right) + (B(1) + 1)(1 - 2) \\ 10 \amp= 10 - B + 2 \\ B \amp= 2 \end{align*}

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Now we know the values of \(A\text{,}\) \(B\text{,}\) and \(C\text{,}\) we obtain

\begin{equation*} \frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)} = \frac{5}{x - 2} + \frac{2x - 2}{x^2 - 2x + 3} \end{equation*}

This implies that

\begin{align*} \int\frac{7x^2 - 16x + 19}{(x - 2)\left(x^2 - 2x + 3\right)}\, dx \amp= \int\left( \frac{5}{x - 2} + \frac{2x - 2}{x^2 - 2x + 3} \right)\, dx \\ \amp= 5\int\frac{1}{x - 2}\, dx + \int \frac{2x - 2}{x^2 - 2x + 3}\, dx \\ \amp= 5\ln|x - 2| + \ln\left(x^2 - 2x + 3\right) + C \end{align*}

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Note: The second integral on the right-hand side can be evaluated quickly using \(u\)-sub with \(u = x^2 - 2x + 3\text{.}\)

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To sum up, we looked at three different cases of the form of the partial fraction decomposition depending on the factors of the denominator. These three cases have different setups. Let’s summarize them into a nice little list below:

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Let \(R(x) = \frac{f(x)}{q(x)}\) be a proper rational expression, where \(q(x)\) is factored completely over the real numbers.

A factor of \(x - r\) in the denominator requires the fraction \(\dfrac{A}{x - r}\) in the setup.
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A repeated linear factor \((x - r)^m\) with \(m > 1\) in the denominator requires

\begin{equation*} \frac{A_1}{x - r} + \frac{A_2}{(x - r)^2} + \frac{A_3}{(x - r)^3} + \cdots + \frac{A_m}{(x - r)^m} \end{equation*}

in the setup.

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An irreducible quadratic factor of \(ax^2 + bx + c\) in the denominator requires the fraction \(\dfrac{Ax + B}{ax^2 + bx + c}\) in the setup.
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We will just need to apply these rules as you see fit based on the factors of the denominator to set up the form of the partial fraction decomposition.

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But Richard... What if there is an irreducible cubic factor in the denominator? Or an irreducible fourth-degree factor in the denominator? Does the pattern of setting the numerator to be one degree less than the denominator also hold?

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Richard’s response: It is impossible to have an irreducible polynomial of degree three and above over the real numbers. That is, if you see a polynomial of degree three or above, you can always factor it. Factor the polynomial into smaller pieces of factors and apply the rule(s).

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But Richard... What if we are given an improper fraction? Can we do something about it first since the method of partial fraction decomposition doesn’t work for improper fractions.

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Richard’s response: You can. You can always obtain a proper fraction out of an improper fraction by doing long division. Then you can apply the method of partial fraction decomposition on the proper fraction.

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Worksheet Some Exercises for this section

I included some practice problems that cover some main concepts in this section. You don’t need to turn it in, but I highly encourage you to work on this with your classmates. I may take problems here to be your in-class practice problems, homework problems, and/or exam problems. Reach out to Richard for help if you get stuck or have any questions.

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I will only include the final answer to some of the problems for you to check your result. If you want to check your work, talk to Richard and he is happy to discuss the process with you.

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Exercise Group.

Set up the form of the partial fraction decomposition for the following rational expressions.

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1.

\(\dfrac{x - 9}{x^2 - 3x - 18}\)

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Solution.

\begin{equation*} \frac{x - 9}{x^2 - 3x - 18} = \frac{A}{x - 6} + \frac{B}{x + 3} \end{equation*}

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2.

\(\dfrac{x^2 - 3x}{x^3 - 3x^2 - 4x}\)

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Solution.

\begin{equation*} \dfrac{x^2 - 3x}{x^3 - 3x^2 - 4x} = \frac{A}{x} + \frac{B}{x - 4} + \frac{C}{x + 1} \end{equation*}

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3.

\(\dfrac{1}{(x + 1)^2\left(x^2 + 2\right)}\)

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Solution.

\begin{equation*} \dfrac{1}{(x + 1)^2\left(x^2 + 2\right)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{Cx + D}{x^2 + 2} \end{equation*}

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Exercise Group.

Evaluate the following integrals.

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4.

\(\displaystyle \int \frac{2x - 1}{x^2 - 5x + 6}\, dx\)

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Solution.

\begin{equation*} 5\ln|x - 3| - 3\ln|x - 2| + C \end{equation*}

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5.

\(\displaystyle \int \frac{dx}{(x - 2)(x - 3)(x + 2)}\, dx\)

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Solution.

\begin{equation*} \frac{1}{20}\ln|x + 2| - \frac{1}{4}\ln|x - 2| + \frac{1}{5}\ln|x - 3| + C \end{equation*}

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6.

\(\displaystyle \int_{-1}^1 \frac{x}{(x + 3)^2}\, dx\)

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Solution.

\begin{equation*} \ln(2) - \frac{3}{4} \approx -0.0568528194401 \end{equation*}

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Exercise Group.

Evaluate the following integrals by first making a substitution to convert the integrand into a rational expression.

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7.

\(\displaystyle \int \frac{e^x\, dx}{\left(e^x - 1\right)\left(e^x + 2\right)}\)

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Solution.

\begin{equation*} \frac{1}{3}\ln\left|\frac{e^x - 1}{e^x + 2}\right| + C \end{equation*}

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8.

\(\displaystyle \int \frac{e^x\, dx}{e^{2x} - e^x}\)

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Solution.

\begin{equation*} \ln\left|e^x - 1\right| - x + C \end{equation*}

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9.

\(\displaystyle \int \frac{\sqrt{x}\, dx}{x - 1}\)

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Solution.

\begin{equation*} 2\sqrt{x} - \ln\left(\sqrt{x} + 1\right) + \ln\left|\sqrt{x} - 1\right| + C \end{equation*}

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Section 7.5 The Method of Partial Fractions

Objectives

Investigation 7.5.1.

Subsection Prerequisite: Proper Rational Expression

Subsection Partial Fraction Decomposition: Distinct Linear Factors

Example 7.5.1.

Example 7.5.2.

Example 7.5.3.

Subsection Partial Fraction Decomposition: Repeated Factors

Example 7.5.4.

Subsection Partial Fraction Decomposition: Irreducible Quadratic Factors

Example 7.5.5.

Worksheet Some Exercises for this section

Exercise Group.

1.

2.

3.

Exercise Group.

4.

5.

6.

Exercise Group.

7.

8.

9.

Worksheet Some Exercises for this section