Integrals Involving
Quadratics
To this point we’ve
seen quite a few integrals that involve quadratics. A couple of examples are,
We also saw that integrals involving 
,
and could be done with a trig substitution.
Notice however that all of these integrals were missing an x term.
They all consist of a quadratic term and a constant.
Some integrals involving general quadratics are easy enough
to do. For instance, the following integral
can be done with a quick substitution.
Some integrals with quadratics can be done with partial
fractions. For instance,
Unfortunately, these methods won’t work on a lot of
integrals. A simple substitution will
only work if the numerator is a constant multiple of the derivative of the
denominator and partial fractions will only work if the denominator can be
factored.
This section is how to deal with integrals involving
quadratics when the techniques that we’ve looked at to this point simply won’t
work.
Back in the Trig
Substitution section we saw how to deal with square roots that had a
general quadratic in them. Let’s take a
quick look at another one like that since the idea involved in doing that kind of
integral is exactly what we are going to need for the other integrals in this
section.
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Example 1 Evaluate
the following integral.
Solution
Recall from the Trig Substitution section that in order to
do a trig substitution here we first needed to complete the square on the
quadratic. This gives,
After completing the square the integral becomes,
Upon doing this we can identify the trig substitution that
we need. Here it is,
Recall that since we are doing an indefinite integral we
can drop the absolute value bars.
Using this substitution the integral becomes,
We can finish the integral out with the following right
triangle.
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So, by completing the square we were able to take an
integral that had a general quadratic in it and convert it into a form that
allowed us to use a known integration technique.
Let’s do a quick review of completing the square before
proceeding. Here is the general
completing the square formula that we’ll use.
This will always take a general quadratic and write it in
terms of a squared term and a constant term.
Recall as well that in order to do this we must have a
coefficient of one in front of the x2. If not we’ll need to factor out the
coefficient before completing the square.
In other words,
Now, let’s see how completing the square can be used to do
integrals that we aren’t able to do at this point.
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Example 2 Evaluate
the following integral.
Solution
Okay, this doesn’t factor so partial fractions just won’t
work on this. Likewise, since the
numerator is just “1” we can’t use the substitution . So, let’s see what happens if we complete
the square on the denominator.
With this the integral is,
Now this may not seem like all that great of a
change. However, notice that we can
now use the following substitution.
and the integral is now,
We can now see that this is an inverse tangent! So, using the formula from above we get,
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Example 3 Evaluate
the following integral.
Solution
This example is a little different from the previous
one. In this case we do have an x in the numerator however the
numerator still isn’t a multiple of the derivative of the denominator and so
a simple Calculus I substitution won’t work.
So, let’s again complete the square on the denominator and
see what we get,
Upon completing the square the integral becomes,
At this point we can use the same type of substitution
that we did in the previous example.
The only real difference is that we’ll need to make sure that we plug
the substitution back into the numerator as well.
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So, in general when dealing with an integral in the form,
Here we are going to assume that the denominator doesn’t
factor and the numerator isn’t a constant multiple of the derivative of the
denominator. In these cases we complete
the square on the denominator and then do a substitution that will yield an
inverse tangent and/or a logarithm depending on the exact form of the numerator.
Let’s now take a look at a couple of integrals that are in
the same general form as (1) except the denominator
will also be raised to a power. In other
words, let’s look at integrals in the form,
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(2)
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Example 4 Evaluate
the following integral.
Solution
For the most part this integral will work the same as the
previous two with one exception that will occur down the road. So, let’s start by completing the square on
the quadratic in the denominator.
The integral is then,
Now, we will use the same substitution that we’ve used to
this point in the previous two examples.
Now, here is where the differences start cropping up. The first integral can be done with the
substitution and isn’t too difficult. The second integral however, can’t be done
with the substitution used on the first integral and it isn’t an inverse
tangent.
It turns out that a trig substitution will work nicely on
the second integral and it will be the same as we did when we had square roots in the problem.
With these two substitutions the integrals become,
Okay, at this point we’ve got two options for the
remaining integral. We can either use
the ideas we learned in the section about
integrals involving trig integrals or we could use the following formula.
Let’s use this formula to do the integral.
Next, let’s use the following right triangle to get this
back to x’s.
The cosine integral is then,
All told then the original integral is,
It’s a long and messy answer, but there it is.
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Example 5 Evaluate
the following integral.
Solution
As with the other problems we’ll first complete the square
on the denominator.
The integral is,
Now, let’s do the substitution.
and the integral is now,
In the first integral we’ll use the substitution
and in the second integral we’ll use the following trig
substitution
Using these substitutions the integral becomes,
We’ll need the following right triangle to finish this
integral out.
So, going back to x’s
the integral becomes,
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Often the following formula is needed when using the trig
substitution that we used in the previous example.
Note that we’ll only need the two trig substitutions that we
used here. The third trig substitution
that we used will not be needed here.