In this article, we will discuss how to approach partial fractions and provide some practice questions to help you test your new skills.
Throughout the Extension 2 course, you will come across many different types of functions to integrate, some of which are easier than others. Partial fraction decomposition is a new method to allow for the integration of rational functions that can’t be integrated directly using the other techniques learned.
Students should be familiar with functions and polynomials including long polynomial division, as well as integration by substitution, and integration resulting in natural logarithms or inverse tan.
A rational function is that which can be written as the ratio of two polynomials.
Partial fraction decomposition involves splitting these expressions up into simpler components, with the goal of an easier integration.
An example of such a function is the following:
\(\frac{5x+2}{x^2+3x+6}\)
The first step in any partial fraction problem is to completely factorise the denominator, in this case resulting in
\(\frac{5x+2}{(x-2)(x-3)}\)
From here we can express the function as a sum of 2 simpler functions, for now leaving the numerators as unknown constants.
The main part of partial fraction decomposition is determining their values.
\(\frac{A}{x-2}+\frac{B}{x-3}\)
It is very important to recognise when this method is applicable.
Partial fraction decomposition can only be done when the degree of the numerator is less than the degree of the denominator!
For example, the function \(\frac{x^2-1}{x^2+3x+6}\) is not able to be decomposed, as the numerator and denominator are both second order.
If you find that the degree of the numerator is larger than the denominator, you can simplify the denominator using polynomial division and then use the partial fractions method.
One method we can use to determine these values is similar to coefficient matching with polynomials, which you would have done previously.
We have the decomposed form of the function, and we have the original form of the function, so we can equate these two!
Multiplying out the denominator of the original expression leaves us with a nice equation without any fractions.
To perform coefficient matching, the RHS should then be factorised in terms of powers of x.
\(5x+2=A(x-3)+B(x-2)\\
⇒ \ 5x+2=x(A+B)+(-3A-2B)\)
This is now a very simple problem to solve! By equating the coefficients of x and the constant, we have a system of 2 equations that can be solved simultaneously.
\begin{align*}
A+B&=5\\
-3A-3B&=2\\
\end{align*}
Solving these yields A = -12 and B = 17, so after substituting these back into the original expression we have:
\(\frac{5x+2}{x^2+3x+6} = \frac{-12}{x-2} + \frac{17}{x-3}\)
We now have two very simple functions to integrate!
The other method that can be used involves substituting in the roots of the denominator.
Starting from after we multiply out the denominator, we have:
\(5x+2=A(x-3)+B(x-2)\)
It’s important to note that for this method the expression is not factored in terms of powers of x, but left in terms of the constants.
Substituting in the roots of the denominator, in this case x = 3 and x = 2:
| \begin{align*} x &= 3 \ \ case: \ 15+2=B(3-2) \ ⇒ \ B=17\\ x &= 2 \ \ case: \ 10+2=A(2-3) \ ⇒ \ A=-12\\ \end{align*} |
Yielding the same results as before!
Neither of these methods are particularly difficult, but some functions will be easier with one or the other.
Determining which to use comes with experience, and often it will be quicker (or necessary for more difficult functions) to use a combination of both.
The previous example was quite simple, and you may not always come across functions which have only linear factors in the denominator. The method for decomposition is a bit different in this case.
Consider the following function:
\(\frac{x+2}{(x^2+2)(x-1)}\)
The denominator has an \(x^2\) which cannot be factored any further. Rather than having 2 values A and B to solve for, we now have the form:
\(\frac{Ax+B}{x^2+2} + \frac{C}{x-1}\)
This can then be solved using the coefficient and substitution methods.
The method of partial fractions works for factors larger than 2nd order as well, but won’t be seen in the HSC course.
The other case which you’ll need to consider is repeated linear factors in the denominator.
Consider the function
\(\frac{x+2}{(x+2)^2(x-1)}\)
The partial fraction decomposition for this function would be
\(\frac{A}{(x+2)^2}+\frac{B}{x+2}+\frac{C}{x-1}\)
Following on from the first example, you should already see that partial fractions are a very useful tool for simplifying integrations.
\(\int \frac{-12}{x-2}+\frac{17}{x-3} \ dx\\
= 17ln(|x-3|) – 12ln(|x-2|)+c\)
Once you’ve decomposed the function, you simply integrate as normal!
Harder problems such as those with repeated or nonlinear factors may require a substitution for one or more of the expressions.
1. Solve \(\int \frac{3x+1}{x^2-6x+8} \ dx\)
2. Solve \( \frac{x+7}{(x^2+1)(x-1)} \ dx\)
3. Solve \( \frac{x+7}{(x+1)^2(x-1)} \ dx\)
1. \( \frac{1}{2}(7ln(|x-2|)+13ln(|x-4|)+c\)
2. \(-2ln(|x^2+1|)-3tan^{-1}(x)+4ln(|x-1|)+c\)
3. \(-2ln(|x+1|)+\frac{3}{x+1}+2ln(|x-1|)+c\)
At Matrix+, our HSC experts will break down difficult concepts through our structured video lessons, and provide you with comprehensive resources that will introduce you to a variety of different questions. Learn more now.
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