Section 7.4 : Partial Fractions
In this section we are going to take a look at integrals of rational expressions of polynomials and once again let’s start this section out with an integral that we can already do so we can contrast it with the integrals that we’ll be doing in this section.
\[\begin{align*}\int{{\frac{{2x - 1}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{1}{u}\,du}}\hspace{0.25in}{\mbox{using }}u = {x^2} - x - 6\,\,{\mbox{ and }}du = \left( {2x - 1} \right)dx\\ & = \ln \left| {{x^2} - x - 6} \right| + c\end{align*}\]So, if the numerator is the derivative of the denominator (or a constant multiple of the derivative of the denominator) doing this kind of integral is fairly simple. However, often the numerator isn’t the derivative of the denominator (or a constant multiple). For example, consider the following integral.
\[\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}\]In this case the numerator is definitely not the derivative of the denominator nor is it a constant multiple of the derivative of the denominator. Therefore, the simple substitution that we used above won’t work. However, if we notice that the integrand can be broken up as follows,
\[\frac{{3x + 11}}{{{x^2} - x - 6}} = \frac{4}{{x - 3}} - \frac{1}{{x + 2}}\]then the integral is actually quite simple.
\[\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}\]This process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition. Many integrals involving rational expressions can be done if we first do partial fractions on the integrand.
So, let’s do a quick review of partial fractions. We’ll start with a rational expression in the form,
\[f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}}\]where both \(P\left( x \right)\) and \(Q\left( x \right)\) are polynomials and the degree of \(P\left( x \right)\) is smaller than the degree of \(Q\left( x \right)\). Recall that the degree of a polynomial is the largest exponent in the polynomial. Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.
So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.
| Factor in denominator |
Term in partial fraction decomposition |
|---|---|
| \( \displaystyle ax + b\) | \( \displaystyle \frac{A}{{ax + b}}\) |
| \( \displaystyle {\left( {ax + b} \right)^k}\) | \( \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} + \cdots + \frac{{{A_k}}}{{{{\left( {ax + b} \right)}^k}}}\), \(k = 1,2,3, \ldots \) |
| \( \displaystyle a{x^2} + bx + c\) | \( \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}}\) |
| \( \displaystyle {\left( {a{x^2} + bx + c} \right)^k}\) | \( \displaystyle \frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots + \frac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}\), \(k = 1,2,3, \ldots \) |
Notice that the first and third cases are really special cases of the second and fourth cases respectively.
There are several methods for determining the coefficients for each term and we will go over each of those in the following examples.
Let’s start the examples by doing the integral above.
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
\[\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{A}{{x - 3}} + \frac{B}{{x + 2}}\]The next step is to actually add the right side back up.
\[\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{{A\left( {x + 2} \right) + B\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\]Now, we need to choose \(A\) and \(B\) so that the numerators of these two are equal for every \(x\). To do this we’ll need to set the numerators equal.
\[3x + 11 = A\left( {x + 2} \right) + B\left( {x - 3} \right)\]Note that in most problems we will go straight from the general form of the decomposition to this step and not bother with actually adding the terms back up. The only point to adding the terms is to get the numerator and we can get that without actually writing down the results of the addition.
At this point we have one of two ways to proceed. One way will always work but is often more work. The other, while it won’t always work, is often quicker when it does work. In this case both will work and so we’ll use the quicker way for this example. We’ll take a look at the other method in a later example.
What we’re going to do here is to notice that the numerators must be equal for any x that we would choose to use. In particular the numerators must be equal for \(x = - 2\) and \(x = 3\). So, let’s plug these in and see what we get.
\[\begin{align*}x & = - 2 : & \hspace{0.5in}5 & = A\left( 0 \right) + B\left( { - 5} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}B & = - 1\\ x & = 3 \,\,\,\,: & \hspace{0.5in}20 & = A\left( 5 \right) + B\left( 0 \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}A & = 4\end{align*}\]So, by carefully picking the \(x\)’s we got the unknown constants to quickly drop out. Note that these are the values we claimed they would be above.
At this point there really isn’t a whole lot to do other than the integral.
\[\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = \int{{\frac{4}{{x - 3}}\,dx}} - \int{{\frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}\]Recall that to do this integral we first split it up into two integrals and then used the substitutions,
\[u = x - 3\hspace{0.5in}v = x + 2\]on the integrals to get the final answer.
Before moving onto the next example a couple of quick notes are in order here. First, many of the integrals in partial fractions problems come down to the type of integral seen above. Make sure that you can do those integrals.
There is also another integral that often shows up in these kinds of problems so we may as well give the formula for it here since we are already on the subject.
\[\int{{\frac{1}{{{x^2} + {a^2}}}\,dx}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c\]It will be an example or two before we use this so don’t forget about it.
Now, let’s work some more examples.
We won’t be putting as much detail into this solution as we did in the previous example. The first thing is to factor the denominator and get the form of the partial fraction decomposition.
\[\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}\]The next step is to set numerators equal. If you need to actually add the right side together to get the numerator for that side then you should do so, however, it will definitely make the problem quicker if you can do the addition in your head to get,
\[{x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)\]As with the previous example it looks like we can just pick a few values of \(x\) and find the constants so let’s do that.
\[\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}\]Note that unlike the first example most of the coefficients here are fractions. That is not unusual so don’t get excited about it when it happens.
Now, let’s do the integral.
\[\begin{align*}\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}} & = \int{{ - \frac{1}{x} + \frac{{\frac{1}{2}}}{{x + 2}} + \frac{{\frac{5}{2}}}{{3x - 2}}\,dx}}\\ & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {x + 2} \right| + \frac{5}{6}\ln \left| {3x - 2} \right| + c\end{align*}\]Again, as noted above, integrals that generate natural logarithms are very common in these problems so make sure you can do them. Also, you were able to correctly do the last integral right? The coefficient of \(\frac{5}{6}\) is correct. Make sure that you do the substitution required for the term properly.
This time the denominator is already factored so let’s just jump right to the partial fraction decomposition.
\[\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\, = \frac{A}{{x - 4}} + \frac{B}{{{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3}}\]Setting numerators gives,
\[{x^2} - 29x + 5 = A\left( {x - 4} \right)\left( {{x^2} + 3} \right) + B\left( {{x^2} + 3} \right) + \left( {Cx + D} \right){\left( {x - 4} \right)^2}\]In this case we aren’t going to be able to just pick values of \(x\) that will give us all the constants. Therefore, we will need to work this the second (and often longer) way. The first step is to multiply out the right side and collect all the like terms together. Doing this gives,
\[{x^2} - 29x + 5 = \left( {A + C} \right){x^3} + \left( { - 4A + B - 8C + D} \right){x^2} + \left( {3A + 16C - 8D} \right)x - 12A + 3B + 16D\]Now we need to choose \(A\), \(B\), \(C\), and \(D\) so that these two are equal. In other words, we will need to set the coefficients of like powers of \(x\) equal. This will give a system of equations that can be solved.
\[\left. \begin{align*}{x^3} & :\hspace{0.25in} & A + C & = 0\\ {x^2} & :\hspace{0.25in} & - 4A + B - 8C + D & = 1\\ {x^1} & :\hspace{0.25in} & 3A + 16C - 8D & = - 29\\ {x^0} & :\hspace{0.25in} & - 12A + 3B + 16D & = 5\end{align*} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 1,\,B = - 5,\,C = - 1,\,D = 2\]Note that we used \({x^0}\) to represent the constants. Also note that these systems can often be quite large and have a fair amount of work involved in solving them. The best way to deal with these is to use some form of computer aided solving techniques.
Now, let’s take a look at the integral.
\[\begin{align*}\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}} & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} + \frac{{ - x + 2}}{{{x^2} + 3}}\,dx}}\\ & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} - \frac{x}{{{x^2} + 3}}\, + \frac{2}{{{x^2} + 3}}\,dx}}\\ & = \ln \left| {x - 4} \right| + \frac{5}{{x - 4}} - \frac{1}{2}\ln \left| {{x^2} + 3} \right| + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + c\end{align*}\]In order to take care of the third term we needed to split it up into two separate terms. Once we’ve done this we can do all the integrals in the problem. The first two use the substitution \(u = x - 4\), the third uses the substitution \(v = {x^2} + 3\) and the fourth term uses the formula given above for inverse tangents.
Let’s first get the general form of the partial fraction decomposition.
\[\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\, = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 4} \right)}^2}}}\]Now, set numerators equal, expand the right side and collect like terms.
\[\begin{align*}{x^3} + 10{x^2} + 3x + 36 & = A{\left( {{x^2} + 4} \right)^2} + \left( {Bx + C} \right)\left( {x - 1} \right)\left( {{x^2} + 4} \right) + \left( {Dx + E} \right)\left( {x - 1} \right)\\ & = \left( {A + B} \right){x^4} + \left( {C - B} \right){x^3} + \left( {8A + 4B - C + D} \right){x^2} + \\ & \hspace{0.5in}\hspace{0.25in}\left( { - 4B + 4C - D + E} \right)x + 16A - 4C - E\end{align*}\]Setting coefficient equal gives the following system.
\[\left. \begin{align*}{x^4} & :\hspace{0.25in} & A + B & = 0\\ {x^3} & :\hspace{0.25in} & C - B & = 1\\ {x^2} & : \hspace{0.25in} & 8A + 4B - C + D & = 10\\ {x^1} & : \hspace{0.25in} & - 4B + 4C - D + E & = 3\\ {x^0} & :\hspace{0.25in} & 16A - 4C - E & = 36\end{align*} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 2,\,B = - 2,\,C = - 1,\,D = 1,\,E = 0\]Don’t get excited if some of the coefficients end up being zero. It happens on occasion.
Here’s the integral.
\[\begin{align*}\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}} & = \int{{\frac{2}{{x - 1}} + \frac{{ - 2x - 1}}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = \int{{\frac{2}{{x - 1}} - \frac{{2x}}{{{x^2} + 4}} - \frac{1}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = 2\ln \left| {x - 1} \right| - \ln \left| {{x^2} + 4} \right| - \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{1}{2}\frac{1}{{{x^2} + 4}} + c\end{align*}\]To this point we’ve only looked at rational expressions where the degree of the numerator was strictly less that the degree of the denominator. Of course, not all rational expressions will fit into this form and so we need to take a look at a couple of examples where this isn’t the case.
So, in this case the degree of the numerator is 4 and the degree of the denominator is 3. Therefore, partial fractions can’t be done on this rational expression.
To fix this up we’ll need to do long division on this to get it into a form that we can deal with. Here is the work for that.
\[\require{enclose} \begin{align*} &\,\,\, x - 2\\ {x^3} - 3{x^2} & \enclose{longdiv}{{x^4} - 5{x^3} + 6{x^2} - 18} \\ - & \underline {\left( {{x^4} - 3{x^3}} \right)\hspace{0.9in}} \\ & \hspace{0.3in} - 2{x^3} + 6{x^2} - 18\\ & \underline { \hspace{0.08in}-\left( { - 2{x^3} + 6{x^2}} \right) \hspace{0.4in}} \\ & \hspace{1.35in} - 18\end{align*}\]So, from the long division we see that,
\[\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\, = x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\]and the integral becomes,
\[\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\\ & = \int{{x - 2\,dx}} - \int{{\frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\end{align*}\]The first integral we can do easily enough and the second integral is now in a form that allows us to do partial fractions. So, let’s get the general form of the partial fractions for the second integrand.
\[\frac{{18}}{{{x^2}\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 3}}\]Setting numerators equal gives us,
\[18 = Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}\]Now, there is a variation of the method we used in the first couple of examples that will work here. There are a couple of values of \(x\) that will allow us to quickly get two of the three constants, but there is no value of \(x\) that will just hand us the third.
What we’ll do in this example is pick \(x\)’s to get the two constants that we can easily get and then we’ll just pick another value of \(x\) that will be easy to work with (i.e. it won’t give large/messy numbers anywhere) and then we’ll use the fact that we also know the other two constants to find the third.
\[\begin{align*}x & = 0 : & \hspace{0.25in} 18 & = B\left( { - 3} \right) & \hspace{0.15in}\Rightarrow \hspace{0.25in}B & = - 6\\ x & = 3 : & \hspace{0.25in} 18 & = C\left( 9 \right) & \hspace{0.15in} \Rightarrow \hspace{0.25in}C & = 2\\ x & = 1 : & 18 & = A\left( { - 2} \right) + B\left( { - 2} \right) + C = - 2A + 14 & \hspace{0.15in} \Rightarrow \hspace{0.25in}A & = - 2\end{align*}\]The integral is then,
\[\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2\,dx}} - \int{{ - \frac{2}{x} - \frac{6}{{{x^2}}} + \frac{2}{{x - 3}}\,dx}}\\ & = \frac{1}{2}{x^2} - 2x + 2\ln \left| x \right| - \frac{6}{x} - 2\ln \left| {x - 3} \right| + c\end{align*}\]In the previous example there were actually two different ways of dealing with the \({x^2}\) in the denominator. One is to treat it as a quadratic which would give the following term in the decomposition
\[\frac{{Ax + B}}{{{x^2}}}\]and the other is to treat it as a linear term in the following way,
\[{x^2} = {\left( {x - 0} \right)^2}\]which gives the following two terms in the decomposition,
\[\frac{A}{x} + \frac{B}{{{x^2}}}\]We used the second way of thinking about it in our example. Notice however that the two will give identical partial fraction decompositions. So, why talk about this? Simple. This will work for \({x^2}\), but what about \({x^3}\) or \({x^4}\)? In these cases, we really will need to use the second way of thinking about these kinds of terms.
\[{x^3}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}}\hspace{0.5in}{x^4}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{D}{{{x^4}}}\]Let’s take a look at one more example.
In this case the numerator and denominator have the same degree. As with the last example we’ll need to do long division to get this into the correct form. We’ll leave the details of that to you to check.
\[\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} = \int{{1 + \frac{1}{{{x^2} - 1}}\,dx}} = \int{{dx}} + \int{{\frac{1}{{{x^2} - 1}}\,dx}}\]So, we’ll need to partial fraction the second integral. Here’s the decomposition.
\[\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}}\]Setting numerator equal gives,
\[1 = A\left( {x + 1} \right) + B\left( {x - 1} \right)\]Picking value of \(x\) gives us the following coefficients.
\[\begin{align*}x & = - 1 : & \hspace{0.25in} 1 & = B\left( { - 2} \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}B & = - \frac{1}{2}\\ x & = 1 \,\,\,\, : & \hspace{0.25in}1 & = A\left( 2 \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}A & = \frac{1}{2}\end{align*}\]The integral is then,
\[\begin{align*}\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} & = \int{{dx}} + \int{{\frac{{\frac{1}{2}}}{{x - 1}} - \frac{{\frac{1}{2}}}{{x + 1}}\,dx}}\\ & = x + \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + c\end{align*}\]