Section 7.4: Partial Fractions

In this section, we are going to take a look at integrals of rational expressions of polynomials. Let’s begin by reviewing a familiar integral to show why partial fractions can be so helpful.

Example 1

Consider the integral $ \int \frac{2x - 1}{x^2 - x - 6} \, dx. $ Notice that if we set $ u = x^2 - x - 6 ,$ then $ du = (2x - 1)\,dx .$ The integral becomes:

        $
        \int \frac{2x - 1}{x^2 - x - 6}\,dx 
        = \int \frac{1}{u}\,du
        = \ln\bigl|x^2 - x - 6\bigr| + C
        $
      

Here, the numerator was exactly the derivative of the denominator, making the integral straightforward. But that's not always the case.

Often, the numerator of a rational function is not a simple constant multiple of the derivative of its denominator. In these cases, the integral is more involved. Let’s consider another example:

Example 2

Now, look at $ \int \frac{3x + 11}{x^2 - x - 6}\, dx. $ The numerator here is clearly not the derivative (nor a constant multiple of it) of the denominator. However, we might notice that:

        $
        \frac{3x + 11}{x^2 - x - 6}
        = \frac{4}{x - 3}
        - \frac{1}{x + 2}
        $
      

With this decomposition, the integral is straightforward:

        $
        \int \frac{3x + 11}{x^2 - x - 6}\, dx
        = \int \left( \frac{4}{x - 3} - \frac{1}{x + 2} \right) dx
        = 4 \ln|x - 3| - \ln|x + 2| + C
        $
      

Definition: The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original expression is called partial fraction decomposition. Many integrals involving rational expressions become much more manageable once we decompose the integrand into partial fractions.

A Quick Review of Partial Fractions

We start with a rational expression in the form $ f(x) = \frac{P(x)}{Q(x)} , $ where both $P(x)$ and $Q(x)$ are polynomials, and the degree of $P(x)$ is strictly less than the degree of $Q(x)$. Partial fraction decomposition applies only if the numerator’s degree is strictly smaller than the denominator’s degree, so that is our first requirement.

Once we confirm this requirement, we factor the denominator as completely as possible. Then, for each factor, we use the following table to determine the term(s) we need in the partial fraction decomposition.

Factor in denominator	Term in partial fraction decomposition
$ ax + b $	$ \frac{A}{ax + b} $
$ (ax + b)^k $	$ \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \dots + \frac{A_k}{(ax + b)^k} $ $( k = 1, 2, 3, \ldots )$
$ ax^2 + bx + c $	$ \frac{Ax + B}{ax^2 + bx + c} $
$ (ax^2 + bx + c)^k $	$ \frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \dots + \frac{A_k x + B_k}{(ax^2 + bx + c)^k} $ $( k = 1, 2, 3, \ldots )$

Notice that the first and third cases listed above are just special cases of the second and fourth, respectively. In practice, you’ll determine the coefficients $$ A, B, A_1, B_1, \dots $$ by multiplying both sides of the decomposition by the original denominator and matching terms. We’ll see detailed examples of this process in future sections.

Example 3

Evaluate the following integral:

        $ \int \frac{2x+5}{x^2 - x - 2}\,dx. $
      

First, we factor the denominator: $ x^2 - x - 2 = (x - 2)(x + 1). $ Hence we set up our partial fraction decomposition as:

        $ \frac{2x + 5}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}. $
      

Combining the right-hand side over the common denominator gives:

        $ \frac{A}{x - 2} + \frac{B}{x + 1}
          = \frac{A(x + 1) + B(x - 2)}{(x - 2)(x + 1)}. $
      

Therefore, we need: $ 2x + 5 = A(x + 1) + B(x - 2). $ Choose convenient values of $x$ to find $A$ and $B$. For instance, set $x = 2$:

        $ 2(2) + 5 = A(2 + 1) + B(0) \implies 4 + 5 = 3A \implies A = 3. $
      

Next, set $x = -1$:

        $ 2(-1) + 5 = A(0) + B(-1 - 2) 
          \implies -2 + 5 = -3B \implies 3 = -3B \implies B = -1. $
      

So the decomposition is:

        $ \frac{2x + 5}{(x - 2)(x + 1)} = \frac{3}{x - 2} - \frac{1}{x + 1}. $
      

Now the integral is straightforward:

        $ \int \frac{2x+5}{x^2 - x - 2} \, dx
          = \int \left(\frac{3}{x - 2} - \frac{1}{x + 1}\right) dx
          = 3 \ln|x - 2| - \ln|x + 1| + C. $
      

Example 4

Evaluate the following integral:

        $ \int \frac{5x + 1}{x^2 + 2x - 3}\,dx. $
      

First, factor the denominator: $ x^2 + 2x - 3 = (x + 3)(x - 1). $ Our partial fraction form is:

        $ \frac{5x + 1}{(x + 3)(x - 1)} = \frac{A}{x + 3} + \frac{B}{x - 1}. $
      

Combine over the common denominator:

        $ \frac{A}{x + 3} + \frac{B}{x - 1}
          = \frac{A(x - 1) + B(x + 3)}{(x + 3)(x - 1)}. $
      

We need: $ 5x + 1 = A(x - 1) + B(x + 3). $ Let's pick $x = 1$:

        $ 5(1) + 1 = A(0) + B(1 + 3) 
          \implies 6 = 4B \implies B = \frac{3}{2}. $
      

Next, pick $x = -3$:

        $ 5(-3) + 1 = A(-3 - 1) + B(0) 
          \implies -15 + 1 = -4A \implies -14 = -4A \implies A = \frac{7}{2}. $
      

Thus:

        $ \frac{5x + 1}{(x + 3)(x - 1)} 
          = \frac{\frac{7}{2}}{x + 3} + \frac{\frac{3}{2}}{x - 1}
          = \frac{7}{2} \cdot \frac{1}{x + 3} + \frac{3}{2} \cdot \frac{1}{x - 1}. $
      

Hence the integral becomes:

        $ \int \frac{5x + 1}{x^2 + 2x - 3}\,dx
          = \int \left(\frac{7}{2} \cdot \frac{1}{x + 3} 
            + \frac{3}{2} \cdot \frac{1}{x - 1}\right) dx
          = \frac{7}{2} \ln|x + 3| + \frac{3}{2} \ln|x - 1| + C. $
      

Example 5

Evaluate the following integral:

        $ \int \frac{4x + 1}{2x^2 - x - 1}\,dx. $
      

Factor the denominator: $ 2x^2 - x - 1 = (2x + 1)(x - 1). $ So we write:

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

Combine into a single fraction:

        $ \frac{A}{2x + 1} + \frac{B}{x - 1}
          = \frac{A(x - 1) + B(2x + 1)}{(2x + 1)(x - 1)}. $
      

We need: $ 4x + 1 = A(x - 1) + B(2x + 1). $ Let's plug in values. If $x = 1$:

        $ 4(1) + 1 = A(0) + B(2(1) + 1) 
          \implies 5 = 3B \implies B = \frac{5}{3}. $
      

If $x = -\frac{1}{2}$:

        $ 4\left(-\frac{1}{2}\right) + 1 = A\left(-\frac{1}{2} - 1\right) + B\cdot 0 
          \implies -2 + 1 = A\left(-\frac{3}{2}\right) 
          \implies -1 = -\frac{3}{2}A 
          \implies A = \frac{2}{3}. $
      

Thus the decomposition is:

        $ \frac{4x + 1}{(2x + 1)(x - 1)} 
          = \frac{\frac{2}{3}}{2x + 1} + \frac{\frac{5}{3}}{x - 1}
          = \frac{2}{3} \cdot \frac{1}{2x + 1} 
            + \frac{5}{3} \cdot \frac{1}{x - 1}. $
      

Now integrate term by term:

        $ \int \frac{4x + 1}{2x^2 - x - 1}\,dx
          = \int \left(\frac{2}{3} \cdot \frac{1}{2x + 1} 
            + \frac{5}{3} \cdot \frac{1}{x - 1}\right) dx.
          = \frac{2}{3} \cdot \frac{1}{2} \ln|2x + 1| 
          + \frac{5}{3} \ln|x - 1| + C.
          = \frac{1}{3} \ln|2x + 1| + \frac{5}{3} \ln|x - 1| + C. $
      

Example 6

Evaluate the following integral:

        $ \int \frac{3x + 13}{x^2 + 2x - 24}\,dx. $
      

Factor the denominator: $ x^2 + 2x - 24 = (x + 6)(x - 4). $ Set up partial fractions:

        $ \frac{3x + 13}{(x + 6)(x - 4)}
          = \frac{A}{x + 6} + \frac{B}{x - 4}. $
      

Combine over the common denominator:

        $ \frac{A}{x + 6} + \frac{B}{x - 4}
          = \frac{A(x - 4) + B(x + 6)}{(x + 6)(x - 4)}. $
      

So we need: $ 3x + 13 = A(x - 4) + B(x + 6). $ Pick $x = 4$:

        $ 3(4) + 13 = A(0) + B(4 + 6)
          \implies 12 + 13 = 10B \implies 25 = 10B \implies B = \frac{5}{2}. $
      

Pick $x = -6$:

        $ 3(-6) + 13 = A(-6 - 4) + B(0)
          \implies -18 + 13 = -10A \implies -5 = -10A \implies A = \frac{1}{2}. $
      

Thus:

        $ \frac{3x + 13}{(x + 6)(x - 4)} 
          = \frac{\frac{1}{2}}{x + 6} + \frac{\frac{5}{2}}{x - 4}
          = \frac{1}{2} \cdot \frac{1}{x + 6} 
            + \frac{5}{2} \cdot \frac{1}{x - 4}. $
      

The integral is then:

        $ \int \frac{3x + 13}{x^2 + 2x - 24}\,dx
          = \int \left(\frac{1}{2} \cdot \frac{1}{x + 6} 
            + \frac{5}{2} \cdot \frac{1}{x - 4}\right) dx
          = \frac{1}{2} \ln|x + 6| + \frac{5}{2} \ln|x - 4| + C. $
      

Example 7

Evaluate the following integral:

        $ \int \frac{7x - 5}{(x - 3)(x + 2)}\,dx. $
      

The denominator is already factored: $(x - 3)(x + 2)$. We assume the partial fractions form:

        $ \frac{7x - 5}{(x - 3)(x + 2)}
          = \frac{A}{x - 3} + \frac{B}{x + 2}. $
      

Combine over the common denominator:

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

Thus we need: $ 7x - 5 = A(x + 2) + B(x - 3). $ Let's pick $x = 3$:

        $ 7(3) - 5 = A(3 + 2) + B(0) 
          \implies 21 - 5 = 5A \implies 16 = 5A \implies A = \frac{16}{5}. $
      

Next, pick $x = -2$:

        $ 7(-2) - 5 = A(0) + B(-2 - 3) 
          \implies -14 - 5 = -5B \implies -19 = -5B \implies B = \frac{19}{5}. $
      

So the decomposition is:

        $ \frac{7x - 5}{(x - 3)(x + 2)}
          = \frac{\frac{16}{5}}{x - 3} + \frac{\frac{19}{5}}{x + 2}
          = \frac{16}{5} \cdot \frac{1}{x - 3}
            + \frac{19}{5} \cdot \frac{1}{x + 2}. $
      

Finally, integrate term by term:

        $ \int \frac{7x - 5}{(x - 3)(x + 2)}\,dx
          = \int \left( \frac{16}{5}\cdot\frac{1}{x - 3} 
            + \frac{19}{5}\cdot\frac{1}{x + 2} \right) dx
          = \frac{16}{5}\ln|x - 3| + \frac{19}{5}\ln|x + 2| + C. $
      

Example 8

Evaluate the following integral:

        $ \int \frac{x^2 + 3x - 5}{x^3 - x^2 - 2x} \, dx. $
      

First, factor the denominator: $ x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x - 2)(x + 1). $ Then we set up the partial fraction decomposition:

        $ \frac{x^2 + 3x - 5}{x(x - 2)(x + 1)} 
          = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}. $
      

Combine into a single fraction:

        $ \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}
          = \frac{A(x - 2)(x + 1) + B(x)(x + 1) + C(x)(x - 2)}{x(x - 2)(x + 1)}. $
      

Thus, we need: $ x^2 + 3x - 5 = A(x - 2)(x + 1) + B\bigl[x(x + 1)\bigr] + C\bigl[x(x - 2)\bigr]. $

A quick way is to choose values for $x$ that simplify terms:

        $ x = 0: \quad (0)^2 + 3(0) - 5 = -5 = A(0 - 2)(0 + 1) = -2A \implies A = \frac{5}{2}. $
        
        $ x = 2: \quad (2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5 
                  = B\bigl[2(2 + 1)\bigr] = 2 \cdot 3B = 6B \implies B = \frac{5}{6}. $
        
        $ x = -1: \quad (-1)^2 + 3(-1) - 5 = 1 - 3 - 5 = -7 
                   = C\bigl[(-1)(-1 - 2)\bigr] = C\bigl[-1 \cdot -3 \bigr] = 3C \implies C = -\frac{7}{3}. $

So the decomposition is:

        $ \frac{x^2 + 3x - 5}{x(x - 2)(x + 1)}
          = \frac{\frac{5}{2}}{x} + \frac{\frac{5}{6}}{x - 2} - \frac{\frac{7}{3}}{x + 1}. $
      

Therefore, the integral becomes:

        $ \int \frac{x^2 + 3x - 5}{x^3 - x^2 - 2x}\, dx
          = \int \left( \frac{5}{2} \cdot \frac{1}{x} 
            + \frac{5}{6} \cdot \frac{1}{x - 2} 
            - \frac{7}{3} \cdot \frac{1}{x + 1} \right) dx
          = \frac{5}{2} \ln|x| + \frac{5}{6} \ln|x - 2| - \frac{7}{3} \ln|x + 1| + C. $
      

Example 9

Evaluate the following integral:

        $ \int \frac{2x - 3}{(x-1)(x^2 + 4)} \, dx. $
      

The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2 + 4)$. So we assume the partial fraction decomposition:

        $ \frac{2x - 3}{(x-1)(x^2 + 4)}
          = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 4}. $
      

Combine into a single fraction:

        $ \frac{A}{x-1} + \frac{Bx + C}{x^2 + 4}
          = \frac{A(x^2 + 4) + (Bx + C)(x-1)}{(x-1)(x^2 + 4)}. $
      

Hence we need: $ 2x - 3 = A(x^2 + 4) + (Bx + C)(x - 1). $

Let’s pick a convenient value for $x$. If $x = 1$:

        $ 2(1) - 3 = A\bigl(1^2 + 4\bigr) + (B\cdot 1 + C)(0) 
          \implies -1 = A(5) \implies A = -\frac{1}{5}. $
      

For the remaining unknowns, expand and compare coefficients or choose more values. Expanding:

        $ A(x^2 + 4) + (Bx + C)(x - 1)
          = -\frac{1}{5} (x^2 + 4) + Bx^2 + Cx - Bx - C. $
        
        $ = \left(B - \frac{1}{5}\right)x^2 + (C - B)x + \left(-\frac{4}{5} - C\right). $

We want this to match $2x - 3$. So the $x^2$ coefficient must be 0:

        $ B - \frac{1}{5} = 0 \implies B = \frac{1}{5}, $
      

the $x$ coefficient is 2:

        $ C - B = 2 \implies C - \frac{1}{5} = 2 \implies C = \frac{11}{5}, $
      

and the constant term is -3:

        $ -\frac{4}{5} - C = -3 \implies -\frac{4}{5} - \frac{11}{5} = -3 \implies -\frac{15}{5} = -3, $
      

which checks out. So:

        $ \frac{2x - 3}{(x-1)(x^2 + 4)}
          = -\frac{1}{5}\cdot \frac{1}{x-1} + \frac{\frac{1}{5}x + \frac{11}{5}}{x^2 + 4}. $
      

Now integrate term by term:

        $ \int \frac{2x - 3}{(x-1)(x^2 + 4)} \, dx
          = -\frac{1}{5}\int \frac{1}{x-1}\,dx 
            + \int \frac{\frac{1}{5}x + \frac{11}{5}}{x^2 + 4}\,dx. $
      

Break the second integral into two parts:

        $ -\frac{1}{5}\ln|x-1|
          + \frac{1}{5}\int \frac{x}{x^2 + 4}\,dx
          + \frac{11}{5}\int \frac{1}{x^2 + 4}\,dx. $
      

We know: $ \int \frac{x}{x^2+4}\,dx = \frac{1}{2}\ln(x^2+4), $ and $ \int \frac{1}{x^2+4}\,dx = \frac{1}{2}\tan^{-1}\Bigl(\frac{x}{2}\Bigr). $

        $ = -\frac{1}{5}\ln|x-1| 
          + \frac{1}{10}\ln|x^2 + 4| 
          + \frac{11}{10}\tan^{-1}\Bigl(\frac{x}{2}\Bigr) + C. $
      

Simplify to get the final result:

        $ \int \frac{2x - 3}{(x-1)(x^2 + 4)} \, dx
          = -\frac{1}{5}\ln|x-1| 
            + \frac{1}{10}\ln(x^2 + 4) 
            + \frac{11}{10}\tan^{-1}\Bigl(\frac{x}{2}\Bigr) 
          + C. $
      

Example 10

Evaluate the following integral:

        $ \int \frac{x^3 - 2x + 4}{x^2(x - 3)}\,dx. $
      

Factor the denominator (already partially factored): $ x^2(x - 3). $ We write:

        $ \frac{x^3 - 2x + 4}{x^2 (x - 3)}
          = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 3}. $
      

Since the numerator degree is not less than the denominator degree, we first do polynomial division.

        $ \frac{x^3 - 2x + 4}{x^2(x-3)}
          = \frac{x^3 - 2x + 4}{x^3 - 3x^2}. $
      

Performing long division gives quotient $1$ and remainder $3x^2 - 2x + 4$, so:

        $ \frac{x^3 - 2x + 4}{x^2 (x-3)} 
          = 1 + \frac{3x^2 - 2x + 4}{x^2 (x-3)}. $
      

Now decompose the remainder part:

        $ \frac{3x^2 - 2x + 4}{x^2(x-3)}
          = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3}. $
      

Matching coefficients gives:

        $ A + C = 3, \quad -3A + B = -2, \quad -3B = 4. $
      

Solving: $ B = -\frac{4}{3}, \; A = \frac{2}{9}, \; C = \frac{25}{9}. $

        $ \frac{x^3 - 2x + 4}{x^2(x - 3)}
          = 1 + \frac{\frac{2}{9}}{x} - \frac{\frac{4}{3}}{x^2} + \frac{\frac{25}{9}}{x-3}. $
      

Now integrate:

        $ \int \frac{x^3 - 2x + 4}{x^2(x - 3)}\,dx
          = x + \frac{2}{9}\ln|x| + \frac{4}{3x} + \frac{25}{9}\ln|x - 3| + C. $
      

Example 11

Evaluate the following integral:

        $ \int \frac{3x + 5}{(x^2 + 1)(x + 2)} \, dx. $
      

The denominator has an irreducible quadratic $x^2 + 1$ and a linear factor $x+2$. So we set up:

        $ \frac{3x + 5}{(x^2 + 1)(x + 2)}
          = \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 2}. $
      

Matching coefficients yields: $ A = \frac{1}{5}, \; B = \frac{13}{5}, \; C = -\frac{1}{5}. $

        $ \frac{3x + 5}{(x^2 + 1)(x + 2)} 
          = \frac{\frac{1}{5}x + \frac{13}{5}}{x^2 + 1} - \frac{\frac{1}{5}}{x + 2}. $
      

Integrate term by term:

        $ \int \frac{3x + 5}{(x^2 + 1)(x + 2)} \, dx
          = \frac{1}{10}\ln(x^2+1) + \frac{13}{5}\tan^{-1}(x) - \frac{1}{5}\ln|x+2| + C. $
      

Example 12

Evaluate the following integral:

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

The denominator factors as $(x-1)^2 (x+1)$. So we set up:

        $ \frac{5x^2 - 1}{(x-1)^2 (x+1)} 
          = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}. $
      

Matching coefficients gives: $ A = 4, \; B = 2, \; C = 1. $

        $ \frac{5x^2 - 1}{(x - 1)^2 (x+1)} 
          = \frac{4}{x-1} + \frac{2}{(x-1)^2} + \frac{1}{x+1}. $
      

Now integrate:

        $ \int \frac{5x^2 - 1}{(x - 1)^2 (x+1)}\, dx
          = 4 \ln|x-1| - \frac{2}{x-1} + \ln|x+1| + C. $
      

Partial Fractions Calculus II | Step-by-Step Guide

Section 7.4: Partial Fractions

Example 1

Example 2

A Quick Review of Partial Fractions

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12