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## www.mathcentre.ac.uk/.../9.6%20Integration%20using%20algebraic%20fractions.mp4

• 0:01 - 0:05
In this video, we're going to
have a look at how we can
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integrate algebraic
fractions. The sorts of
• 0:07 - 0:09
fractions that we're going to
integrate and like these
• 0:09 - 0:09
here.
• 0:11 - 0:14
Now, superficially, they will
look very similar, but there are
• 0:14 - 0:18
important differences which I'd
like to point out when you come
• 0:18 - 0:21
to tackle a problem of
integrating a fraction like
• 0:21 - 0:24
this, it's important that you
can look for certain features,
• 0:24 - 0:26
for example, in this first
• 0:26 - 0:31
example. In the denominator we
have what we call 2 linear
• 0:31 - 0:34
factors. These two linear
factors are two different linear
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factors. When I say linear
factors, I mean there's no X
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squared, no ex cubes, nothing
like that in it. These are just
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of the form a X Plus B linear
factors. A constant times X plus
• 0:48 - 0:51
another constant. So with two
linear factors here.
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This example is also got linear
factors in clearly X Plus One is
• 0:56 - 1:01
a linear factor. X minus one is
a linear factor, but the fact
• 1:01 - 1:04
that I've got an X minus one
squared means that we've really
• 1:04 - 1:08
got X minus one times X minus 1
two linear factors within here.
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So we call this an example of a
repeated linear factor.
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Linear factors, repeated linear
factors. Over here we've got a
• 1:19 - 1:24
quadratic. Now this particular
quadratic will not factorize and
• 1:24 - 1:29
because it won't factorize, we
call it an irreducible quadratic
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factor. This final example here
has also got a quadratic factor
• 1:34 - 1:37
in the denominator, but unlike
the previous one, this one will
• 1:37 - 1:41
in fact factorize because X
squared minus four is actually
• 1:41 - 1:45
the difference of two squares
and we can write this as X minus
• 1:45 - 1:49
2X plus two. So whilst it might
have originally looked like a
• 1:49 - 1:53
quadratic factor, it was in fact
to linear factors, so that's
• 1:53 - 1:56
what those are one of the
important things we should be
• 1:56 - 1:59
looking for when we come to
integrate quantities like these.
• 1:59 - 2:01
Weather we've got linear
• 2:01 - 2:04
factors. Repeated linear
factors, irreducible quadratic
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factors, or quadratic factors
that will factorize.
• 2:07 - 2:11
Something else which is
important as well, is to examine
• 2:11 - 2:14
the degree of the numerator and
the degree of the denominator in
• 2:14 - 2:17
each of these fractions.
Remember, the degree is the
• 2:17 - 2:21
highest power, so for example in
the denominator of this example
• 2:21 - 2:24
here, if we multiplied it all
out, we actually get the highest
• 2:24 - 2:28
power as three, because when we
multiply the first terms out
• 2:28 - 2:32
will get an X squared, and when
we multiply it with the SEC
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bracket, X Plus, one will end up
with an X cubed. So the degree
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of the denominator there is 3.
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The degree of the numerator is 0
because we can think of this as
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One X to the 0.
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In the first case, we've got an
X to the one here, so the degree
• 2:50 - 2:54
of the numerator there is one,
and if we multiply the brackets
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out, the degree of the of the
denominator will be two. Will
• 2:57 - 3:01
get a quadratic term in here. In
this case, the degree of the
• 3:01 - 3:05
numerator is 0. And in this
case, the degree of the
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denominator is to the highest
power is too. So in all of
• 3:10 - 3:14
these cases, the degree of the
numerator is less than the
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degree of the denominator, and
we call fractions like these
• 3:17 - 3:18
proper fractions.
• 3:20 - 3:24
On the other hand, if we look at
this final example, the degree
• 3:24 - 3:28
of the numerator is 3, whereas
the degree of the denominator is
• 3:28 - 3:32
too. So in this case, the degree
of the numerator is greater than
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the degree of the denominator,
and this is what's called an
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improper fraction. Now when we
start to integrate quantities
• 3:39 - 3:42
like this will need to examine
whether we're dealing with
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proper fractions or improper
fractions, and then, as I said
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before, will need to look at all
the factors in the denominator.
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Will also need to call appan
techniques in the theory of
• 3:53 - 3:57
partial fractions. There is a
video on partial fractions and
• 3:57 - 4:00
you may wish to refer to that if
• 4:00 - 4:04
necessary. If you have a linear
factor in the denominator, this
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will lead to a partial fraction
of this form. A constant over
• 4:08 - 4:09
the linear factor.
• 4:10 - 4:14
If you have a repeated linear
factor in the denominator,
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you'll need two partial
fractions. A constant over
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the factor and a constant
over the factor squared.
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Finally, if you have a quadratic
factor which is irreducible,
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you'll need to write a partial
fraction of the form a constant
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times X plus another constant
over the irreducible quadratic
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factor, so will certainly be
calling upon the techniques of
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partial fractions. Will also
need to call appan. Lots of
• 4:42 - 4:43
techniques and integration.
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I'm just going to mention just
two or three here, which will
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need to use as we proceed
through the examples of 1st.
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Crucial result is the standard
result, which says that if you
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have an integral consisting of a
function in the denominator.
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And it's derivative in the
numerator. Then the result is
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the logarithm of the modulus of
the function in the denominator.
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So for example, if I ask you to
integrate one over X plus one
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with respect to X.
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Then clearly the function in the
denominator is X plus one.
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And its derivative is one which
appears in the numerator. So
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we've an example of this form.
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So the resulting integral is the
logarithm of the modulus of the
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function that was in the
denominator, which is X plus
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one. Plus a constant of
integration, so we will need
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that result very frequently in
the examples which are going to
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follow will also need some
standard results and one of the
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standard results I will call
appan is this one. The integral
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of one over a squared plus X
squared is one over a inverse
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tan of X over a plus C.
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Results like this can be found
in tables of standard integrals.
• 5:57 - 6:01
Finally, we need to integrate
quantities like this and you'll
• 6:01 - 6:05
need to do this probably using
integration by substitution. An
• 6:05 - 6:09
integral like this can be worked
out by making the substitution
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you equals X minus one.
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So that the differential du
is du DX.
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DX, which in this case is du
DX, will be just one.
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So do you is DX and that's
integral. Then will become
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the integral of one over
you, squared du.
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One over you squared is the same
as the integral of you to the
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minus 2D U, which you can solve
by integrating increasing the
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power BI want to give you you to
the minus one over minus one.
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Plus a constant of integration.
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This can be finished off by
changing the you back to the
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original variable X minus one
and that will give us X minus
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one to the minus one over minus
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one plus C. Which is the same as
minus one over X minus one plus
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C, which is the results I have
• 7:05 - 7:09
here. So what I'm saying is that
throughout the rest of this unit
• 7:09 - 7:13
will need to call Appan lots of
different techniques to be able
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to perform the integrals.
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As we shall see.
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Let's look at the
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first example. Suppose we
want to integrate this algebraic
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fraction. 6 / 2
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minus X. X
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+3
DX
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The first thing we do is we look
at the object we've got and try
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to ask ourselves, are we dealing
with a proper or improper
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fraction and what are the
factors in the denominator like?
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Well, if we multiply the power,
the brackets at the bottom will
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find that the highest power of X
is X squared, so the degree of
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the denominator is 2.
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The highest power in the
numerator is one. This is an X
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to the power one, so the degree
of the numerator is one because
• 8:09 - 8:12
the degree of the numerator is
less than the degree of the
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denominator. This is an example
of a proper fraction.
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Both of these factors
in the denominator.
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Are linear factors.
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So we're dealing with a proper
fraction with linear factors.
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The way we proceed is to take
this fraction and express it in
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partial fractions. So I'll start
with the fraction again.
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And express it in the
appropriate form of partial
• 8:49 - 8:52
fractions. Now because it's
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proper. And because we've got
linear factors, the appropriate
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form is to have a constant over
the first linear factor.
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Plus another constant over the
second linear factor.
• 9:08 - 9:12
Our task now is to find values
for the constants A&B.
• 9:12 - 9:16
Once we've done that, will
be able to evaluate this
• 9:16 - 9:18
integral by evaluating these
two separately.
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So to find A and be the first
thing we do is we add these
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two fractions together again.
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Remember that to add 2 fractions
together, we've got to give them
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the same denominator. They've
got to have a common
• 9:35 - 9:39
denominator. The common
denominator is going to be made
• 9:39 - 9:41
up of the two factors. 2 minus
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X&X +3. To write the first term
as an equivalent fraction with
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this denominator, we multiply
top and bottom by X plus three.
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So if we multiply top here by X
+3 and bottom there by X +3.
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Will achieve this fraction.
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And this fraction is equivalent
to the original 1.
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Similarly with the second term.
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To achieve a common denominator
of 2 minus XX +3.
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I need to multiply top and
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bottom here. By two minus X, so
B times 2 minus X and this
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denominator times 2 minus X and
that will give me.
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B2 minus X at the top.
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Now these two fractions
have the same denominator,
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we can add them together
simply by adding the
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numerators together, which
will give us a multiplied
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by X +3.
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Plus B multiplied by
two minus X.
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All divided by the common
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denominator. What we're saying
is that this fraction we
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started with is exactly the
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same. As this quantity here.
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Now the denominators
are already the same.
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So if this is the same as that,
and the denominators are already
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the same, then so too must be
the numerators, so we can equate
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the numerators if we equate the
numerators we can write down X
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equals. AX
+3.
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Plus B2 Minus
X.
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This is the equation that's
going to allow us to calculate
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values for A&B.
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Now we can find values for A&B
in one of two different ways.
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The 1st way that I'm going to
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look at. Is to substitute
specific values in for X.
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Remember that this quantity on
the left is supposed to be equal
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to this on the right for any
value of X at all. So in
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particular, we can choose any
values that we like. That will
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make all this look simpler.
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And what I'm going to do is I'm
going to choose X to have the
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value to. Why would I do that?
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I choose X to have the value
too, because then this second
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term will become zero and have 2
- 2, which is zero. Will lose
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this term. And we'll be able to
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calculate A. So by careful
choice of values for X, we can
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make this look a lot simpler.
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So with X is 2.
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On the left will have two.
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On the right will have 2 +
3, which is 5 times a.
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And this term will vanish.
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This gives me a value for a
straightaway dividing both sides
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by 5. I can write that a is 2/5.
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We need to find B.
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Now a sensible value that will
enable us to find B is to let X
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be minus three whi, is that?
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Well, if X is minus three,
will have minus 3 + 3, which
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is zero. And all of this
first term will vanish.
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And we'll be able to find be so
letting XP minus three will have
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minus three on the left zero
from this term here, and two
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minus minus three, which is 2 +
3, which is 55-B.
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Dividing both sides by 5 will
give us a value for B's
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minus three over 5.
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So now we know a value for a. We
know a value for B.
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And we can then proceed to
evaluate the integral by
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evaluating each of these
• 13:49 - 13:56
separately. Let me write
this down again. We want
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the integral of X divided
by 2 minus XX +3.
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With respect to X.
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We expressed this algebraic
fraction in its partial fraction
• 14:10 - 14:14
in its partial fractions, and we
found that a was 2/5.
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And be was
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minus 3/5. So instead
of integrating this original
• 14:25 - 14:31
fraction, what we're going to do
now is integrate separately the
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two partial fractions.
• 14:33 - 14:38
And will integrate these
separately and will do it like
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this. In the first integral,
we're going to take out the
• 14:42 - 14:47
factor of 2/5. I will be left
with the problem of integrating
• 14:47 - 14:50
one over 2 minus X with respect
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to X. For the second,
we're going to take out
• 14:57 - 15:03
minus 3/5. And integrate
one over X +3 with
• 15:03 - 15:05
respect to X.
• 15:08 - 15:10
So the problem of
integrating this algebraic
• 15:10 - 15:12
fraction has been split
into the problem of
• 15:12 - 15:15
evaluating these two
separate integrals and both
• 15:15 - 15:18
of these are simpler than
the one we started with.
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Let's deal with the second one
first. The second one is a
• 15:23 - 15:26
situation where we've got a
function at the bottom and it's
• 15:26 - 15:30
derivative at the top. Because
we've got X plus three at the
• 15:30 - 15:34
bottom and the derivative of X
+3 is just one which appears at
• 15:34 - 15:38
the top. So this just evaluates
to minus 3/5 the natural
• 15:38 - 15:41
logarithm of the modulus of
what's at the bottom.
• 15:43 - 15:45
We've got the similar situation
here, except if you
• 15:45 - 15:48
differentiate the denominator,
you get minus one because of
• 15:48 - 15:53
this minus X, so we'd really
like a minus one at the top.
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And I can adjust my numerator to
make it minus one, provided that
• 15:58 - 16:01
I counteract that with putting a
minus sign outside there.
• 16:02 - 16:07
So we can write all this as
minus 2/5 the natural logarithm
• 16:07 - 16:13
of the modulus of 2 minus X. And
of course we need a constant of
• 16:13 - 16:15
integration at the very end.
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So that's the result of
integrating X over 2 minus XX +3
• 16:21 - 16:22
and the problems finished.
• 16:23 - 16:28
What I'd like to do is just go
back a page and just show you an
• 16:28 - 16:32
alternative way of calculating
values for the constants A&B in
• 16:32 - 16:36
the partial fractions, and I
want us to return to this
• 16:36 - 16:40
equation here that we use to
find A&B. Let me write that
• 16:40 - 16:41
equation down again.
• 16:41 - 16:48
X is equal to
a X +3.
• 16:48 - 16:51
Plus B2 Minus
• 16:51 - 16:58
X. What I'm going
to do is I'm going to
• 16:58 - 17:01
start by removing the brackets.
• 17:02 - 17:05
Will have a multiplied by X
• 17:05 - 17:10
AX. A Times 3
which is 3A.
• 17:12 - 17:15
B times two or two B.
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And be times minus X or
minus BX.
• 17:21 - 17:25
And then what I'm going to do
is, I'm going to collect similar
• 17:25 - 17:28
terms together so you see Ivan
Axe here and minus BX there.
• 17:29 - 17:34
So altogether I have a minus B,
lots of X.
• 17:37 - 17:42
And we've got 3A Plus
2B here.
• 17:45 - 17:48
We now use this equation to
equate coefficients on both
• 17:48 - 17:52
sides. What do we mean by that?
Well, what we do is we ask
• 17:52 - 17:57
ourselves how many X terms do we
have on the left and match that
• 17:57 - 18:02
with the number of X terms that
we have on the right. So you
• 18:02 - 18:07
see, on the left hand side here,
if we look at just the ex terms,
• 18:07 - 18:13
there's 1X. On the right we've
got a minus B, lots of X.
• 18:13 - 18:18
So we've equated the
coefficients of X on both sides.
• 18:19 - 18:23
We can also look at constant
terms on both sides. You see the
• 18:23 - 18:25
three A plus 2B is a constant.
• 18:26 - 18:30
There are no constant terms on
the left, so if we just look at
• 18:30 - 18:32
constants, there are none on the
• 18:32 - 18:37
left. And on the right there's
3A Plus 2B.
• 18:37 - 18:41
And you'll see what we have.
Here are two simultaneous
• 18:41 - 18:44
equations for A&B and if we
solve these equations we can
• 18:44 - 18:49
find values for A&B. Let me call
that equation one and that one
• 18:49 - 18:53
equation two. What I'm going to
do is I'm going to multiply
• 18:53 - 18:58
equation one by two so that will
end up with two be so that we
• 18:58 - 19:01
will be able to add these
together to eliminate the bees.
• 19:01 - 19:06
So if I take equation one and I
multiply it by two, I'll get 2
• 19:06 - 19:14
ones or two. 2A minus 2B.
Let's call that equation 3.
• 19:14 - 19:18
If we add equations two and
• 19:18 - 19:25
three together. We've got 0 + 2,
which is 2, three, 8 + 2 A which
• 19:25 - 19:31
is 5A and two be added to minus
2B cancels out, so two is 5. In
• 19:31 - 19:36
other words, A is 2 over 5,
which is the value we had
• 19:36 - 19:37
earlier on for A.
• 19:38 - 19:43
We can then take this value for
A and substitute it in either of
• 19:43 - 19:46
these equations and obtain a
value for be. So, for example,
• 19:46 - 19:48
if we substitute in the first
• 19:48 - 19:55
equation. Will find that one
equals a, is 2/5 minus B.
• 19:56 - 20:02
Rearranging this B is equal to
2/5 - 1.
• 20:03 - 20:09
And 2/5 - 1 is minus 3/5 the
same value as we got before.
• 20:10 - 20:14
So we've seen two ways of
finding the values of the
• 20:14 - 20:18
constants A&B. We can substitute
specific values for X or we can
• 20:18 - 20:19
equate coefficients on both
• 20:19 - 20:23
sides. Often will need to use a
mix of the two methods in order
• 20:23 - 20:25
to find all the constants in a
• 20:25 - 20:31
given problem. Let's have
a look at
• 20:31 - 20:34
a definite integral.
• 20:35 - 20:39
Suppose we want to find the
integral from X is one to access
• 20:39 - 20:47
2. Of three divided by
XX plus one with respect to
• 20:47 - 20:53
X. As before, we examine this
integrand and ask ourselves, is
• 20:53 - 20:55
this a proper or improper
• 20:55 - 20:59
fraction? Well, the degree of
the denominator is too, because
• 20:59 - 21:03
when we multiply this out, the
highest power of X will be 2.
• 21:04 - 21:09
The degree of the numerator is
zero, with really 3X to the zero
• 21:09 - 21:13
here, so this is an example of a
• 21:13 - 21:18
proper fraction. On both of
these factors are linear
• 21:18 - 21:24
factors. So as before, I'm going
to express the integrand.
• 21:25 - 21:27
As the sum of its partial
fractions. So let's do that
• 21:27 - 21:31
first of all. 3 divided by XX
• 21:31 - 21:36
plus one. The appropriate form
of partial fractions.
• 21:37 - 21:43
Are constant. Over the
first linear factor.
• 21:43 - 21:48
Plus another constant over the
second linear factor.
• 21:49 - 21:53
And our job now is to try to
find values for A&B.
• 21:54 - 22:00
We do this by adding these
together as we did before,
• 22:00 - 22:05
common denominator XX plus one
in both cases.
• 22:07 - 22:11
To write a over X as an
equivalent fraction with
• 22:11 - 22:14
this denominator will
need to multiply top and
• 22:14 - 22:16
bottom by X plus one.
• 22:18 - 22:24
To write B over X plus one with
this denominator will need to
• 22:24 - 22:27
multiply top and bottom by X.
• 22:27 - 22:33
So now we've given these two
fractions a common denominator,
• 22:33 - 22:38
and we add the fractions
together by adding the
• 22:38 - 22:43
numerators. I'm putting the
result over the common
• 22:43 - 22:51
denominator. So 3 divided by XX
plus One is equal to all this.
• 22:53 - 22:55
The denominators are already the
• 22:55 - 23:00
same. So we can equate the
numerators that gives us the
• 23:00 - 23:05
equation 3 equals a X plus one
plus BX. And this is the
• 23:05 - 23:10
equation we can use to try to
find values for A&B.
• 23:11 - 23:14
We could equate coefficients, or
we can substitute specific
• 23:14 - 23:17
values for X and what I'm going
to do is I'm going to substitute
• 23:17 - 23:21
the value X is not and the
reason why I'm picking X is not
• 23:21 - 23:23
is because I recognize straight
away that's going to.
• 23:24 - 23:27
Kill off this last term here
that'll have gone and will be
• 23:27 - 23:28
able to just find a value for A.
• 23:29 - 23:35
So we substitute X is not on the
left, will still have 3.
• 23:36 - 23:42
And on the right we've got not
plus one which is one 1A.
• 23:42 - 23:44
Be times not is not so that
• 23:44 - 23:48
goes. So In other words, we've
got a value for A and a is 3.
• 23:50 - 23:54
Another sensible value to
substitute is X equals minus
• 23:54 - 23:57
one. Why is that a sensible
• 23:57 - 24:00
value? Well, that's a sensible
value, because if we put X is
• 24:00 - 24:04
minus one in minus one plus one
is zero and will lose this first
• 24:04 - 24:06
term with the A in and will now
• 24:06 - 24:10
be. So putting X is minus one
will have 3.
• 24:10 - 24:14
This will become zero and will
have be times minus one which is
• 24:14 - 24:20
minus B. So this tells us that B
is actually minus three.
• 24:20 - 24:26
So now we know the value of a is
going to be 3 and B is going to
• 24:26 - 24:31
be minus three. And the problem
of performing this integration
• 24:31 - 24:34
can be solved by integrating
these two terms separately.
• 24:36 - 24:41
Let's do that now. I'll write
these these terms down again.
• 24:41 - 24:47
We're integrating three over XX
plus one with respect to X.
• 24:48 - 24:52
And we've expressed already this
as its partial fractions, and
• 24:52 - 24:57
found that we're integrating
three over X minus three over X,
• 24:57 - 25:02
plus one with respect to X, and
this was a definite integral. It
• 25:02 - 25:06
had limits on, and the limits
will one and two.
• 25:07 - 25:10
So now we use partial
fractions to change this
• 25:10 - 25:13
algebraic fraction into
these two simple integrals.
• 25:14 - 25:16
Now, these are
straightforward to finish
• 25:16 - 25:20
because the integral of
three over X is just three
• 25:20 - 25:22
natural logarithm of the
modulus of X.
• 25:26 - 25:27
The integral of three over X
• 25:27 - 25:33
plus one. Is 3 natural logarithm
of the modulus of X plus one?
• 25:33 - 25:36
And there's a minus sign
in the middle from that.
• 25:37 - 25:40
This is a definite
• 25:40 - 25:44
integral. So we have square
brackets and we write the limits
• 25:44 - 25:46
on the right hand side.
• 25:46 - 25:49
The problem is nearly
finished. All we have to do
• 25:49 - 25:50
is substitute the limits in.
• 25:51 - 25:57
Upper limit first when X is
2, will have three natural
• 25:57 - 25:58
log of two.
• 25:59 - 26:03
When X is 2 in here will have
• 26:03 - 26:06
minus three. Natural
logarithm of 2 + 1, which is
• 26:06 - 26:11
3. So that's what we get when
we put the upper limit in.
• 26:12 - 26:17
When we put the lower limit in,
when X is one will have three
• 26:17 - 26:18
natural logarithm of 1.
• 26:19 - 26:23
Minus three natural logarithm of
1 + 1, which is 2. So that's
• 26:23 - 26:28
what we get when we put the
lower limiting. And of course we
• 26:28 - 26:31
want to find the difference of
these two quantities.
• 26:32 - 26:36
Here you'll notice with three
• 26:36 - 26:41
log 2. And over here
there's another three
• 26:41 - 26:44
log, two with A minus and
minus, so we're adding
• 26:44 - 26:48
another three log 2. So
altogether there will be
• 26:48 - 26:49
6 log 2.
• 26:50 - 26:53
That's minus three log 3.
• 26:53 - 26:57
And the logarithm of
• 26:57 - 27:01
1. Is 0 so that banishes.
• 27:01 - 27:04
Now we could leave the answer
like that, although more often
• 27:04 - 27:07
than not would probably use
the laws of logarithms to try
• 27:07 - 27:11
to tighten this up a little
bit and write it in a
• 27:11 - 27:13
different way. You should be
aware that multiplier outside,
• 27:13 - 27:17
like this six, can be put
inside as a power, so we can
• 27:17 - 27:20
write this as logarithm of 2
to the power 6.
• 27:21 - 27:24
Subtract again a multiplier
outside can move inside as a
• 27:24 - 27:28
power so we can write this as
logarithm of 3 to the power 3.
• 27:29 - 27:33
And you'll also be aware from
your loss of logarithms that if
• 27:33 - 27:36
we're finding the difference of
two logarithms, and we can write
• 27:36 - 27:41
that as the logarithm of 2 to
the power 6 / 3 to the power 3.
• 27:41 - 27:44
And that's my final answer.
• 27:45 - 27:52
Let's look at another example in
which the denominator contains a
• 27:52 - 27:57
repeated linear factor. Suppose
we're interested in evaluating
• 27:57 - 28:05
this integral 1 divided by X
minus one all squared X Plus
• 28:05 - 28:12
One, and we want to integrate
that with respect to X.
• 28:13 - 28:15
So again, we have a proper
• 28:15 - 28:19
fraction. And there is a linear
factor here. X plus one, another
• 28:19 - 28:23
linear factor X minus one. But
this is a repeated linear factor
• 28:23 - 28:25
because it occurs twice.
• 28:26 - 28:31
The appropriate form of partial
fractions will be these.
• 28:31 - 28:38
We want to
constant over the
• 28:38 - 28:41
linear factor X
• 28:41 - 28:47
minus one. We want
another constant over the linear
• 28:47 - 28:53
factor repeated X minus one
squared. And finally we need
• 28:53 - 28:58
another constant. See over this
linear factor X plus one.
• 28:58 - 29:02
And our task is before
is to try to find values
• 29:02 - 29:03
for the constants AB&C.
• 29:04 - 29:09
We do that as before, by
expressing each of these over a
• 29:09 - 29:12
common denominator and the
common denominator that we want
• 29:12 - 29:15
is going to be X minus one.
• 29:16 - 29:20
Squared X plus
one.
• 29:21 - 29:25
Now to achieve a common
denominator of X minus one
• 29:25 - 29:29
squared X Plus one will need to
multiply the top and bottom here
• 29:29 - 29:35
by X minus 1X plus one. So we
have a X minus 1X plus one.
• 29:35 - 29:38
To achieve the common
denominator in this case will
• 29:38 - 29:42
need to multiply top and bottom
by X plus one.
• 29:43 - 29:47
So we'll have a BX plus one.
• 29:47 - 29:51
And finally, in this case, to
achieve a denominator of X minus
• 29:51 - 29:56
one squared X Plus one will need
to multiply top and bottom by X
• 29:56 - 29:57
minus 1 squared.
• 29:58 - 30:04
Now this fraction here
is the same as
• 30:04 - 30:07
this fraction here.
• 30:07 - 30:10
Their denominators are already
the same, so we can equate the
• 30:10 - 30:14
numerators. So if we just look
at the numerators will have one
• 30:14 - 30:18
on the left is equal to the top
line here on the right hand
• 30:18 - 30:25
side. AX minus
1X plus one.
• 30:25 - 30:29
Plus B. X plus one.
• 30:30 - 30:34
Plus C. X minus one all
• 30:34 - 30:39
squared. And now we choose some
sensible values for X, so
• 30:39 - 30:42
there's a lot of these terms
will drop away. For example,
• 30:42 - 30:46
supposing we pick X equals 1,
what's the point of picking X
• 30:46 - 30:50
equals one? Well, if we pick
axes one, this first term
• 30:50 - 30:52
vanishes. We lose a.
• 30:52 - 30:56
Also, if we pick X equal to 1,
the last term vanish is because
• 30:56 - 31:00
we have a 1 - 1 which is zero
and will be just left with the
• 31:00 - 31:04
term involving be. So by letting
XP, one will have one.
• 31:05 - 31:08
On the left is equal to 0.
• 31:08 - 31:12
One and one here is 22B.
• 31:12 - 31:19
And the last term vanishes. In
other words, B is equal to 1/2.
• 31:20 - 31:25
What's another sensible value to
pick for X? Well, if we let X
• 31:25 - 31:26
equal minus one.
• 31:27 - 31:31
X being minus one will mean that
this term vanish is minus one
• 31:31 - 31:32
plus One is 0.
• 31:33 - 31:37
This term will vanish minus one,
plus one is zero and will be
• 31:37 - 31:41
able to find see. So I'm going
to let X be minus one.
• 31:42 - 31:45
Will still have the one on the
• 31:45 - 31:48
left. When X is minus one, this
• 31:48 - 31:50
goes. This goes.
• 31:51 - 31:55
And on the right hand side this
term here will have minus 1 - 1
• 31:55 - 32:00
is minus two we square it will
get plus four, so will get plus
• 32:00 - 32:05
4C. In other words, see is going
to be 1/4.
• 32:06 - 32:11
So we've got be. We've got C and
now we need to find a value for
• 32:11 - 32:15
a. Now we can substitute any
other value we like in here, so
• 32:15 - 32:19
I'm actually going to pick X
equals 0. It's a nice simple
• 32:19 - 32:23
value. Effects is zero, will
have one on the left if X is
• 32:23 - 32:26
zero there and there will be
left with minus one.
• 32:27 - 32:32
Minus 1 * 1 is minus 1 - 1
times a is minus a.
• 32:33 - 32:39
Thanks is 0 here. Will just left
with B Times one which is be.
• 32:39 - 32:44
And if X is 0 here will have
minus one squared, which is plus
• 32:44 - 32:50
one plus One Times C Plus C. Now
we already know values for B and
• 32:50 - 32:56
for C, so we substitute these in
will have one is equal to minus
• 32:56 - 33:00
A plus B which is 1/2 plus C
which is 1/4.
• 33:01 - 33:06
So rearranging this will have
that a is equal to.
• 33:07 - 33:12
Well, with a half and a quarter,
that's 3/4 and the one from this
• 33:12 - 33:18
side over the other side is
minus one. 3/4 - 1 is going to
• 33:18 - 33:19
be minus 1/4.
• 33:19 - 33:26
So there we have our values for
a for B and for C, so a being
• 33:26 - 33:29
minus 1/4 will go back in there.
• 33:29 - 33:34
Be being a half will go back in
here and see being a quarter
• 33:34 - 33:38
will go back in there and then
we'll have three separate pieces
• 33:38 - 33:41
of integration to do in order to
complete the problem.
• 33:42 - 33:46
Let me write these all down
again. I'm going to write the
• 33:46 - 33:49
integral out and write these
three separate ones down.
• 33:49 - 33:57
So we had that the integral of
one over X minus one squared X
• 33:57 - 34:03
Plus one, all integrated with
respect to X, is going to equal.
• 34:04 - 34:10
Minus 1/4 the integral of one
over X minus one.
• 34:11 - 34:18
They'll be plus 1/2 integral of
one over X minus 1 squared.
• 34:18 - 34:24
And they'll
be 1/4
• 34:24 - 34:27
over X
• 34:27 - 34:33
plus one.
And we want
• 34:33 - 34:36
to DX in
• 34:36 - 34:42
every term. So we've
used partial fractions to split
• 34:42 - 34:46
this integrand into three
separate terms, and will try and
• 34:46 - 34:50
finish this off. Now. The first
integral straightforward when we
• 34:50 - 34:54
integrate one over X minus one
will end up with just the
• 34:54 - 34:58
natural logarithm of the
denominator, so we'll end up
• 34:58 - 35:01
with minus 1/4 natural
logarithm. The modulus of X
• 35:01 - 35:06
minus one. To integrate this
term, we're integrating one over
• 35:06 - 35:11
X minus one squared. Let me just
do that separately.
• 35:11 - 35:16
To integrate one over X minus
one all squared, we make a
• 35:16 - 35:20
substitution and let you equals
X minus one.
• 35:21 - 35:25
Do you then will equal du DX,
which is just one times DX, so
• 35:25 - 35:28
do you will be the same as DX?
• 35:28 - 35:32
The integral will become the
integral of one over.
• 35:33 - 35:36
X minus one squared
will be you squared.
• 35:37 - 35:38
And DX is D.
• 35:39 - 35:43
Now, this is straightforward to
finish because this is
• 35:43 - 35:45
integrating you to the minus 2.
• 35:46 - 35:49
Increase the power by one
becomes you to the minus 1
• 35:49 - 35:51
divided by the new power.
• 35:51 - 35:53
And add a constant of
integration.
• 35:54 - 35:59
So when we do this, integration
will end up with minus one over
• 35:59 - 36:01
you. Which is minus one over.
• 36:02 - 36:04
X minus one.
• 36:05 - 36:11
And we can put that back into
here now. So the integral half
• 36:11 - 36:16
integral of one over X minus one
squared will be 1/2.
• 36:16 - 36:20
All that we've got down here,
which is minus one over X minus
• 36:20 - 36:25
one. The constant of integration
will add another very end. This
• 36:25 - 36:27
integral is going to be 1/4.
• 36:28 - 36:32
The natural logarithm of the
modulus of X Plus One and then a
• 36:32 - 36:34
single constant for all of that.
• 36:35 - 36:39
Let me just tidy up this little
a little bit. The two logarithm
• 36:39 - 36:43
terms can be combined. We've got
a quarter of this logarithm.
• 36:43 - 36:47
Subtract 1/4 of this logarithm,
so together we've got a quarter
• 36:47 - 36:51
the logarithm. If we use the
laws of logarithms, we've gotta
• 36:51 - 36:55
log subtracted log. So we want
the first term X Plus one
• 36:55 - 36:57
divided by the second term X
• 36:57 - 37:03
minus one. And then this term
can be written as minus 1/2.
• 37:03 - 37:06
One over X minus one.
• 37:06 - 37:10
Is a constant of integration
at the end, and that's the
• 37:10 - 37:11
integration complete.
• 37:12 - 37:18
Now let's look at an example in
which we have an improper
• 37:18 - 37:21
fraction. Suppose we have this
• 37:21 - 37:27
integral. The
degree of
• 37:27 - 37:31
the numerator
• 37:31 - 37:36
is 3.
The degree of the
• 37:36 - 37:38
denominator is 2.
• 37:39 - 37:42
3 being greater than two
means that this is an
• 37:42 - 37:42
improper fraction.
• 37:43 - 37:46
Improper fractions requires
special treatment and the first
• 37:46 - 37:50
thing we do is division
polynomial division. Now, if
• 37:50 - 37:54
you're not happy with Long
Division of polynomials, then I
• 37:54 - 37:57
would suggest that you look
again at the video called
• 37:57 - 38:01
polynomial division. When
examples like this are done very
• 38:01 - 38:05
thoroughly, but what we want to
do is see how many times X minus
• 38:05 - 38:07
X squared minus four will divide
• 38:07 - 38:13
into X cubed. The way we do
this polynomial division is we
• 38:13 - 38:20
say how many times does X
squared divided into X cubed.
• 38:20 - 38:25
That's like asking How many
times X squared will go into X
• 38:25 - 38:29
cubed, and clearly when X
squared is divided into X cubed,
• 38:29 - 38:31
the answer is just X.
• 38:32 - 38:36
So X squared goes into X
cubed X times and we write
• 38:36 - 38:37
the solution down there.
• 38:38 - 38:42
We take what we have just
written down and multiply it by
• 38:42 - 38:46
everything here. So X times X
squared is X cubed.
• 38:46 - 38:50
X times minus four is minus 4X.
• 38:50 - 38:52
And then we subtract.
• 38:53 - 38:55
X cubed minus X cubed vanish is.
• 38:56 - 39:00
With no access here, and we're
subtracting minus 4X, which is
• 39:00 - 39:02
like adding 4X.
• 39:03 - 39:08
This means that when we divide X
squared minus four into X cubed,
• 39:08 - 39:14
we get a whole part X and a
remainder 4X. In other words, X
• 39:14 - 39:18
cubed divided by X squared minus
four can be written as X.
• 39:19 - 39:25
Plus 4X divided by X
squared minus 4.
• 39:25 - 39:30
So in order to tackle this
integration, we've done the long
• 39:30 - 39:35
division and we're left with two
separate integrals to workout.
• 39:36 - 39:38
Now this one is going to be
straightforward. Clearly you're
• 39:38 - 39:41
just integrating X, it's going
to be X squared over 2. This is
• 39:41 - 39:44
a bit more problematic. Let's
have a look at this again.
• 39:44 - 39:48
This is now a proper fraction
where the degree of the
• 39:48 - 39:51
numerator is one with the next
to the one here.
• 39:53 - 39:55
The degree of the denominator is
• 39:55 - 39:57
2. So it's a proper fraction.
• 39:58 - 40:01
It looks as though we've got a
quadratic factor in the
• 40:01 - 40:06
denominator. But in fact, X
squared minus four will
• 40:06 - 40:10
factorize. It's in fact, the
difference of two squares. So we
• 40:10 - 40:16
can write 4X over X squared
minus four as 4X over X minus 2X
• 40:16 - 40:21
plus two. So the quadratic will
actually factorize and you'll
• 40:21 - 40:23
see. Now we've got two linear
• 40:23 - 40:28
factors. We can express this in
partial fractions. Let's do
• 40:28 - 40:35
that. We've got 4X over X, minus
two X +2, and because each of
• 40:35 - 40:40
these are linear factors, the
appropriate form is going to be
• 40:40 - 40:42
a over X minus 2.
• 40:43 - 40:47
Plus B over X +2.
• 40:50 - 40:55
We add these together using a
common dumped common denominator
• 40:55 - 41:03
X minus two X +2 and will
get a X +2 plus BX minus
• 41:03 - 41:09
2. All over the common
denominator, X minus two X +2.
• 41:12 - 41:19
Now the left hand side and the
right hand side of the same. The
• 41:19 - 41:24
denominators are already the
same, so the numerators must be
• 41:24 - 41:28
the same. 4X must equal a X +2
• 41:28 - 41:30
plus B. 6 - 2.
• 41:31 - 41:36
This is the equation that we can
use to find the values for A&B.
• 41:38 - 41:39
Let me write that down again.
• 41:40 - 41:46
We've got 4X is
a X +2.
• 41:46 - 41:51
Plus BX minus
2.
• 41:53 - 41:57
What's a sensible value to put
in for X? Well, if we choose, X
• 41:57 - 41:59
is 2, will lose the second term.
• 42:00 - 42:03
So X is 2 is a good
value to put in here.
• 42:04 - 42:09
Faxes 2 on the left hand side
will have 4 * 2, which is 8.
• 42:10 - 42:15
If X is 2, will have 2 +
2, which is 44A.
• 42:15 - 42:18
And this term will vanish.
• 42:19 - 42:22
So 8 equals 4A.
• 42:22 - 42:27
A will equal 2 and that's the
value for A.
• 42:29 - 42:33
What's another sensible value to
put in for X? Well, if X is
• 42:33 - 42:36
minus two, will have minus 2 +
2, which is zero. Will lose this
• 42:36 - 42:44
term. So if X is
minus two, will have minus 8
• 42:44 - 42:46
on the left.
• 42:46 - 42:48
This first term will vanish.
• 42:48 - 42:54
An ex being minus two here will
have minus 2 - 2 is minus four
• 42:54 - 42:56
so minus 4B.
• 42:56 - 43:01
So be must also be
equal to two.
• 43:01 - 43:06
So now we've got values for A
and for be both equal to two.
• 43:06 - 43:09
Let's take us back and see what
• 43:09 - 43:11
that means. It means that when
• 43:11 - 43:15
we express. This quantity
4X over X squared minus
• 43:15 - 43:18
four in partial fractions
like this, the values of
• 43:18 - 43:20
A&B are both equal to two.
• 43:21 - 43:27
So the integral were working out
is the integral of X +2 over X
• 43:27 - 43:30
minus 2 + 2 over X +2.
• 43:30 - 43:36
So there we
are. We've now
• 43:36 - 43:42
got three separate
integrals to evaluate,
• 43:42 - 43:48
and it's straightforward
to finish this
• 43:48 - 43:55
off. The integral of X
is X squared divided by two.
• 43:56 - 44:02
The integral of two over X minus
2 equals 2 natural logarithm of
• 44:02 - 44:06
the modulus of X minus 2.
• 44:06 - 44:14
And the integral of two over X
+2 is 2 natural logarithm of X
• 44:14 - 44:17
+2 plus a constant of
• 44:17 - 44:21
integration. If we wanted to do,
we could use the laws of
• 44:21 - 44:24
logarithms to combine these log
rhythmic terms here, but I'll
• 44:24 - 44:25
leave the answer like that.
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