Although adding and subtracting fractions with unlike denominators of a variable is generally taught in middle school, performing the reverse operation is a much more advanced topic and is generally, at earliest, found in a second-year/second-semester Calculus course in the mainstream mathematics curriculum. Although the process of partial fraction decomposition itself does not use calculus, it has applications that are useful for Calculus. The AP Calculus BC exam requires partial fraction decomposition to antidifferentiate rational expressions.
The form of the fraction you are trying to decompose will vary, because the degree of the numerator and the denominator will vary. There is another key point to consider when doing these problems: when you are given a rational expression, begin by factoring the denominator; this is very helpful for determining what type of decomposition you should use.
In addition, if the degree of the denominator is less than 2, this technique will fail because the fraction cannot be decomposed any further.
There is a general rule to follow for this situation, described as follows: these expressions will take the general form of
$$\frac{ax + b}{(cx + d)(jx + k)}$$
where \(a, b, c, d, j,\) and \(k\) are constants, and \(x\) is the variable. This form of a rational expression will decompose into two fractions, each with a constant numerator and a linear denominator.
This is one of the most common forms of fractions that would need to be decomposed. Therefore it is one of the most important to understand.
Here is a basic example that employs this technique.
Example 1: Perform partial fraction decomposition on this rational expression:
$$\frac{x - 4}{x^2 - 5x + 6}$$
Solution: The denominator factors to \((x - 4)(x - 2)\)
Therefore the fraction can be broken up as follows:
$$\frac{A}{x - 3} + \frac{B}{x - 2} = \frac{x - 4}{(x - 3)(x - 2)}$$
Multiply both sides of the equation by \((x - 3)(x - 2)\) to eliminate the fractions:
$$A(x - 2) + B(x - 3) = x - 4$$
Let \(x = 2\) to solve for \(B\), as this eliminates \(A\):
$$-B = 2 - 4\Rightarrow$$ $$-B = -2 \Rightarrow$$ $$B = 2$$
Likewise, let \(x = 3\) to solve for \(A\), because it eliminates \(B\):
$$A = 3 - 4 \Rightarrow$$ $$A = -1$$
Plugging in these coefficients gives
$$-\frac{1}{x - 3} + \frac{2}{x - 2}$$
The next example has the same concepts, but the algebra is slightly harder.
Example 2: Perform partial fraction decomposition on this rational expression:
$$\frac{x - 4}{56x^2 + 61x + 15}$$
Solution: We must first factor the denominator of this expression to determine how we are going to decompose this:
$$\frac{x - 4}{(7x + 5)(8x + 3)}$$
Therefore our decomposition will be of the form
$$\frac{x - 4}{(7x + 5)(8x + 3)} = \frac{A}{7x + 5} + \frac{B}{8x + 3}$$
At this point we can begin solving for \(A\) and \(B\). Multiply both sides of this equation by the denominator of the left-hand side; this ensures that all fractions are eliminated:
$$x - 4 = A(8x + 3) + B(7x + 5)$$
To solve for B, let \(x = -\frac{3}{8}\):
$$-\frac{3}{8} - 4 = A(8 \cdot -\frac{3}{8} + 3) + B(7 \cdot -\frac{3}{8} + 5) \Rightarrow$$ $$-\frac{3}{8} - \frac{32}{8} = A(3 - 3) + B(-\frac{21}{8} + \frac{40}{8}) \Rightarrow$$ $$-\frac{35}{8} = 0A + \frac{19B}{8} \Rightarrow$$ $$-35 = 19B \Rightarrow$$ $$B = -\frac{35}{19}$$
To solve for A, let \(x = -\frac{5}{7}\):
$$-\frac{5}{7} - 4 = A(8 \cdot -\frac{5}{7} + 3) + B(7 \cdot -\frac{5}{7} + 5) \Rightarrow$$ $$-\frac{5}{7} - \frac{28}{7} = A(-\frac{40}{7} + \frac{21}{7}) + B(-5 + 5) \Rightarrow$$ $$-\frac{33}{7} = -\frac{19A}{7} + 0B \Rightarrow$$ $$-33 = -19A \Rightarrow$$ $$A = \frac{33}{19}$$
We can substitute these values of \(A\) and \(B\) into our expression:
$$\frac{\frac{33}{19}}{7x + 5} + \frac{\frac{-35}{19}}{8x + 3} = \frac{33}{133x + 95} - \frac{35}{152x + 57}$$
Notice that the x-values we selected were chosen in order to eliminate one variable. This makes it possible to solve for the other.
Interestingly, the fraction can have a numerator of degree 0--there is no variable in it. This type of fraction is less common than the first but also relatively simple to decompose.
A fraction of the form
$$\frac{m}{(ax + b)(cx + d)}$$
can also be decomposed into the form
$$\frac{A}{ax + b} + \frac{B}{cx + d},$$
where \(A, a, B, b, c,\) and \(d\) are constants, and \(x\) is the variable. The process of decomposition is virtually identical to when the numerator is linear, but the algebra is slightly easier.
Here is an example of decomposing a fraction of this form.
Example 3: Decompose this fraction: \(\frac{7}{27x^2 - 15x - 22}\).
Solution: Factoring the denominator of this expression gives
$$\frac{7}{(3x + 2)(9x - 11)}$$
This will be decomposed as follows:
$$\frac{7}{(3x + 2)(9x - 11)} = \frac{A}{3x + 2} + \frac{B}{9x - 11}$$
Multiply to eliminate the fractions with the same intent as in Example 1:
$$7 = A(9x - 11) + B(3x + 2)$$
Let \(x = \frac{11}{9}\) to solve for \(B\):
$$7 = B(3 \cdot \frac{11}{9} + 2) \Rightarrow$$ $$7 = B(\frac{11}{3} + \frac{6}{3}) \Rightarrow$$ $$7 = \frac{17B}{3} \Rightarrow$$ $$21 = 17B \Rightarrow$$ $$B = \frac{21}{17}$$
Likewise, solve for A by letting \(x = -\frac{2}{3}\):
$$7 = A(9 \cdot -\frac{2}{3} - 11) \Rightarrow$$ $$7 = A(-6 - 11) \Rightarrow$$ $$7 = -17A \Rightarrow$$ $$A = -\frac{7}{17}$$
Thus we get the decomposition
$$\frac{-\frac{7}{17}}{3x + 2} + \frac{\frac{21}{17}}{9x - 11} = -\frac{7}{51x + 34} + \frac{21}{153x - 187}$$
A common application of partial fraction decomposition finds its way into calculus courses. Partial fraction decomposition is considered an integration technique because the decomposition yields expressions that are very easy to integrate with common substitutions. Here are two final examples that put this technique into practice in a calculus context.
Example 4: Integrate the following indefinite integral: \(\int{\frac{3}{2x^2 - 7x + 5}dx}\).
Solution:
To integrate this, it is necessary to perform partial fraction decomposition on the integrand first. Factor the denominator to get
$$\frac{3}{(2x - 1)(x - 5)}$$
The decomposition gives this equation for the unknown constants:
$$\frac{3}{(2x - 1)(x - 5)} = \frac{A}{2x - 1} + \frac{B}{x - 5}$$
Eliminate the fractions:
$$3 = A(x - 5) + B(2x - 1)$$
To solve for \(A\), let \(x = \frac{1}{2}\):
$$3 = A(\frac{1}{2} - \frac{10}{2}) \Rightarrow$$ $$3 = -\frac{9A}{2} \Rightarrow$$ $$A = -\frac{2}{3}$$
To solve for \(B\), let \(x = 5\):
$$3 = B(2(5) - 1) \Rightarrow$$ $$3 = 9B \Rightarrow$$ $$B = \frac{1}{3}$$
Therefore the fraction decomposes to
$$-\frac{2}{6x - 3} + \frac{1}{3x - 15}$$
In order to use a logarithmic substitution on both of the two terms, write the terms so that leading coefficients of the denominators are both \(1\):
$$-\frac{1}{3} \cdot \frac{1}{x - \frac{1}{2}} + \frac{1}{3} \cdot \frac{1}{x - 5}$$
For each term, to integrate make a substitution: \(u = x - a \Rightarrow du = dx\) for the constant in the fractions with the variable. This gives us the final expression
$$-\frac{1}{3}\ln|x - \frac{1}{2}| + \frac{1}{3}\ln|x - 5| + C$$
This of course can be checked by differentiation. If the derivative matches the original integrand that also means that the partial fraction decomposition was also performed correctly.
Partial Fraction Decomposition is one of the harder algebraic techniques to learn, and rarely does a student learn this in a high school algebra course, despite the fact that it would fit very nicely into a standard Algebra II class. There is no need to wait until Calculus to introduce this.
Example 5: Integrate this expression: \(\frac{4x + 6}{2x^2 + 20x + 42}\).
Solution: First and foremost, notice that the numerator and denominator have a common factor of \(2\), so remove that:
$$\frac{2x + 3}{x^2 + 10x + 21}$$
Next, factor the denominator:
$$\frac{2x + 3}{(x + 3)(x + 7)}$$
Now define constants \(A\) and \(B\) so that
$$\frac{2x + 3}{(x + 3)(x + 7)} = \frac{A}{x + 3} + \frac{B}{x + 7}$$
Multiply both sides of the equation by \((x + 3)(x + 7)\) to eliminate the fractions:
$$2x + 3 = A(x + 7) + B(x + 3)$$
First, let \(x = -7\) to solve for \(B\):
$$2(-7) + 3 = B(-7 + 3) \Rightarrow$$ $$-14 + 3 = -4B \Rightarrow$$ $$-11 = -4B \Rightarrow$$ $$B = \frac{11}{4}$$
Next, let \(x = -3\) to solve for \(A\):
$$2(-3) + 3 = A(-3 + 7) \Rightarrow$$ $$-3 = 4A \Rightarrow$$ $$A = -\frac{3}{4}$$
Substitute these values in the expression for integration:
$$-\frac{3}{4} \cdot \frac{1}{x + 3} + \frac{11}{4} \cdot \frac{1}{x + 7}$$
With the same method as in the previous example, each piece can be integrated individually. As expected with these types of expressions, the result involves logarithms:
$$-\frac{3}{4} \ln|x + 3| + \frac{11}{4} \ln |x + 7| + C$$
This is a fairly difficult algebraic technique to master. However, many mathematics curricula postpone the introduction of this topic to AP Calculus BC. There is no need to do this, as the algebraic techniques are nothing learned past Algebra II, meaning the technique can be introduced much earlier.