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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.
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.
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,
then the integral is actually quite simple.
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,
where both P(x)
and Q(x) are polynomials and the
degree of P(x) is smaller than the
degree of Q(x). 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.
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.
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Example 1 Evaluate
the following integral.
Solution
The first step is to factor the denominator as much as
possible and get the form of the partial fraction decomposition. Doing this gives,
The next step is to actually add the right side back up.
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.
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 and . So, let’s plug these in and see what we
get.
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.
Recall that to do this integral we first split it up into
two integrals and then used the substitutions,
on the integrals to get the final answer.
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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.
It will be an example or two before we use this so don’t
forget about it.
Now, let’s work some more examples.
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Example 2 Evaluate
the following integral.
Solution
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.
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,
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.
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.
Again, as noted above, integrals that generate natural
logarithms are very common in these problems so make sure you can do them.
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Example 3 Evaluate
the following integral.
Solution
This time the denominator is already factored so let’s
just jump right to the partial fraction decomposition.
Setting numerators gives,
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,
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.
Note that we used x0
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.
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 ,
the third uses the substitution and the fourth term uses the formula given
above for inverse tangents.
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Example 4 Evaluate
the following integral.
Solution
Let’s first get the general form of the partial fraction
decomposition.
Now, set numerators equal, expand the right side and
collect like terms.
Setting coefficient equal gives the following system.
Don’t get excited if some of the coefficients end up being
zero. It happens on occasion.
Here’s the integral.
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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.
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Example 5 Evaluate
the following integral.
Solution
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.
So, from the long division we see that,
and the integral becomes,
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.
Setting numerators equal gives us,
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.
The integral is then,
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