Partial Fraction Expansion is also called Partial Fraction Decomposition.
So what exactly is partial fraction expansion/decomposition? Here is a great video clip explaining, with an example, that it is the opposite of combining fractions.
video by The Organic Chemistry Tutor |
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The main idea of partial fractions is that you start with a rational function that has a polynomial in the denominator. This polynomial can usually be factored, which will allow you to write the fraction as two or more separate fractions. Why would we want to do this? Well, when you get to integration, you will find that there are many integrals that cannot be integrated directly. If you can expand the fraction using this technique, you may then be able to integrate each individual fraction. The other application that you will run across in first year calculus is in infinite series. Sometimes a telescoping series is hidden in a fraction and, using partial fractions, you can then determine if it is a telescoping series.
Caution: When initially learning this technique be wary of teachers or textbooks or friends, who try to show you shortcuts. Yes, shortcuts may work in some cases but learning the basic technique will allow you apply it to a wide variety of problems. Shortcuts will limit what you can do and, honestly, good teachers will give you a situation on an exam that isn't covered by shortcuts. Also, in a later class, you may come across a situation where shortcuts do not work. So you will just be handcuffing yourself if you spend time learning them. This is true with most math topics. So be careful.
As always, we will show you how to use the technique in all cases.
Partial Fraction Expansion Steps
As we present each of these steps, we will work through an example to show you how they work.
Example To Show Steps: Expand the fraction \(\displaystyle{ \frac{1}{x^2+3x+2} }\) using the technique of partial fraction expansion.
Step 1. You first need to get your fraction in a form that you can apply this technique. There are two things that are important. First, the fraction needs to be in the simplest form possible with a factorable polynomial in the denominator. The idea of simplest form possible is something you develop over time by working practice problems.
Second, after you satisfy the first requirement, if you have a polynomial in the numerator as well, the highest power in the numerator must be less than the highest power in the denominator. If it isn't, you need to do polynomial long division to get the correct form.
Most of the time, the problem will be given to you in the correct form. But just be aware of these requirements. [This example is already in the correct form. See the practice problems on the next two pages for problems where you need to do more work.]
Step 2. Completely factor the denominator using whatever techniques you have.
\(x^2+3x+2 = (x+1)(x+2)\)
Step 3. Write the original fraction as separate fractions, with the denominators as the terms you found in step 2 and with numerators with unknown constants.
\( \displaystyle{ \frac{1}{x^2+3x+2} } = \) \( \displaystyle{ \frac{1}{(x+1)(x+2)} = } \) \( \displaystyle{ \frac{A}{x+1} + \frac{B}{x+2} } \)
Step 4. Multiply both sides of the equation by the factored denominator and cancel like terms.
\( \displaystyle{ \frac{1}{(x+1)(x+2)} [(x+1)(x+2)] }\) \( \displaystyle{ = \left[ \frac{A}{x+1} + \frac{B}{x+2} \right] [(x+1)(x+2)] }\)
\( \displaystyle{ 1 = \frac{A}{x+1} [(x+1)(x+2)] + }\) \( \displaystyle{ \frac{B}{x+2} [(x+1)(x+2)] } \)
\(\displaystyle{ 1 = A(x+2) + B(x+1) }\)
Notice what happened here. On the left side, the denominator completely canceled. On the right, the \(x+1\) canceled in the first fraction and the \(x+2\) term canceled in the second fraction.
Step 5. Find the value of constants in the numerators.
The equation we have so far must hold for all values of \(x\). So we can choose whatever values of \(x\) we want to solve for \(A\) and \(B\). We suggest you choose values so that one of the terms cancels, for example, in this problem, we would choose \(x=-1\) (which cancels the term for \(B\)) and then \(x=-2\). Let's see how this works.
Using the last equation above, we have \(1 = A(x+2) + B(x+1)\)
\(x = -1 → 1 = A(-1+2) + B(-1+1) → 1 = A(1) + B(0) → 1 = A\)
\(x = -2 → 1 = A(-2+2) + B(-2+1) → 1 = A(0) + B(-1) → -1 = B\)
Step 6. Final Answer: Clearly show your resulting equations.
\(\displaystyle{ \frac{1}{x^2+3x+2} = \frac{1}{x+1} + \frac{-1}{x+2} }\)
Complications
Okay, so now you are thinking, is that all there is to it? Well, actually, no. We chose a pretty easy problem to show you the basic steps and, in general, those steps do not change from problem to problem. The complication comes in when setting up the original set of fractions, after factoring (step 3). In the above example, we had very simple terms with the highest power of one in both cases. This told us that we just needed constants (\(A\) and \(B\)) in the numerators. This is not always the case. The numerators with variables can be more complicated but not overly so. Let's look at each case here. Once you get the pattern down and the equations set up, all the other steps are pretty much the same.
Keep in mind that the terms we are looking at must be completely factored (in the real number system) as shown in step 2 above.
Alternate Technique To Find Constants
The technique above (shown in step 5) works in most cases. However, sometimes the equations can get quite messy. (Not that they can't get messy with this technique too but sometimes this is less complicated.)
Note: As always, check with your instructor to see what they expect. Sometimes, they may be emphasizing something by requiring a certain technique. Their requirements come before those found on this site. But mostly, trust them. They may know what they are doing.
One reservation we have with the technique shown above that you may have thought about is the comment that the equation must hold for all values of \(x\). However, if you look closely, the original fraction is not defined at the points that we are sometimes plugging in for \(x\) since the denominator is zero there. (Go back to the example above if this is not clear.) However, plugging in those values works, although why it works is not clear. Therefore, to avoid this situation, we present this technique that, honestly, is clearer and involves less algebra.
You may ask why didn't just show this technique above. Well, we have found that many, if not most, teachers teach that technique. So we presented it up front. In our opinion, this alternate technique is preferred.
Alternate Technique
This technique involves matching coefficients of the variables. Let's use the example above to see how it works. We start out with \(1 = A(x+2) + B(x+1)\), which is the result from step 4. The variable involved is \(x\) and the constants we are trying to find are \(A\) and \(B\). First, we multiply out the right side and combine common terms depending on the variable \(x\) .
\(\begin{array}{rcl}
1 & = & A(x+2) + B(x+1) \\
& = & Ax + 2A + Bx + B \\
& = & Ax + Bx + 2A + B \\
& = & (A+B)x + (2A+B)
\end{array}\)
Okay, so, if we write out explicitly the terms on the left, so that we can see them clearly, we have the equation
\(0x+1 = (A+B)x + (2A+B)\)
Now, we equate the cofficients of the terms to get the two equations
\( 0 = A+B \) and \( 1 = 2A+B \)
Now, we have several ways we can solve these, including substitution, matrices and adding/subtracting the equations. We will use the last technique and subtract the first equation from the second.
\(\begin{array}{rcrcl}
2A & + & B & = & 1 \\
-(~~A & + & B & = & 0~~) \\
-A & - & B & = & 0 \\
---&---&---&---&--- \\
A & & & = & 1
\end{array}\)
Substituting \(A=1\) into the equation \(A + B = 0\) gives us \(B = -1\).
These are the same values we obtained from the above technique. Okay, your next logical step is to learn the details on how to handle linear factors, single and repeating. Go to the Linear Partial Fractions page, read it and work those practice problems.
Really UNDERSTAND Precalculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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