WEBVTT

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In this video, we're going to
have a look at how we can

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integrate algebraic
fractions. The sorts of

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fractions that we're going to
integrate and like these

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here.

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Now, superficially, they will
look very similar, but there are

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important differences which I'd
like to point out when you come

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to tackle a problem of
integrating a fraction like

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this, it's important that you
can look for certain features,

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for example, in this first

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example. In the denominator we
have what we call 2 linear

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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

00:00:38.090 --> 00:00:42.578
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

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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

00:00:55.965 --> 00:01:00.720
a linear factor. X minus one is
a linear factor, but the fact

00:01:00.720 --> 00:01:04.440
that I've got an X minus one
squared means that we've really

00:01:04.440 --> 00:01:08.470
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

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quadratic. Now this particular
quadratic will not factorize and

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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

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in the denominator, but unlike
the previous one, this one will

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in fact factorize because X
squared minus four is actually

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the difference of two squares
and we can write this as X minus

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2X plus two. So whilst it might
have originally looked like a

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quadratic factor, it was in fact
to linear factors, so that's

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what those are one of the
important things we should be

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looking for when we come to
integrate quantities like these.

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Weather we've got linear

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factors. Repeated linear
factors, irreducible quadratic

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factors, or quadratic factors
that will factorize.

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Something else which is
important as well, is to examine

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the degree of the numerator and
the degree of the denominator in

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each of these fractions.
Remember, the degree is the

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highest power, so for example in
the denominator of this example

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here, if we multiplied it all
out, we actually get the highest

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power as three, because when we
multiply the first terms out

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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

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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

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get a quadratic term in here. In
this case, the degree of the

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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

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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

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proper fractions.

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On the other hand, if we look at
this final example, the degree

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of the numerator is 3, whereas
the degree of the denominator is

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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

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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

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partial fractions. There is a
video on partial fractions and

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you may wish to refer to that if

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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

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the linear factor.

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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

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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.

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Finally, we need to integrate
quantities like this and you'll

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need to do this probably using
integration by substitution. An

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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

00:07:04.891 --> 00:07:09.384
here. So what I'm saying is that
throughout the rest of this unit

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will need to call Appan lots of
different techniques to be able

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to perform the integrals.

00:07:14.540 --> 00:07:16.020
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

00:07:45.696 --> 00:07:48.156
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

00:08:04.318 --> 00:08:08.530
to the power one, so the degree
of the numerator is one because

00:08:08.530 --> 00:08:12.418
the degree of the numerator is
less than the degree of the

00:08:12.418 --> 00:08:16.460
denominator. This is an example
of a proper fraction.

00:08:17.040 --> 00:08:22.745
Both of these factors
in the denominator.

00:08:23.380 --> 00:08:24.658
Are linear factors.

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So we're dealing with a proper
fraction with linear factors.

00:08:31.820 --> 00:08:37.787
The way we proceed is to take
this fraction and express it in

00:08:37.787 --> 00:08:41.918
partial fractions. So I'll start
with the fraction again.

00:08:42.770 --> 00:08:49.043
And express it in the
appropriate form of partial

00:08:49.043 --> 00:08:51.831
fractions. Now because it's

00:08:51.831 --> 00:08:56.206
proper. And because we've got
linear factors, the appropriate

00:08:56.206 --> 00:09:00.573
form is to have a constant over
the first linear factor.

00:09:01.560 --> 00:09:06.376
Plus another constant over the
second linear factor.

00:09:07.580 --> 00:09:11.529
Our task now is to find values
for the constants A&B.

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Once we've done that, will
be able to evaluate this

00:09:15.980 --> 00:09:18.236
integral by evaluating these
two separately.

00:09:19.480 --> 00:09:24.145
So to find A and be the first
thing we do is we add these

00:09:24.145 --> 00:09:25.389
two fractions together again.

00:09:26.440 --> 00:09:31.276
Remember that to add 2 fractions
together, we've got to give them

00:09:31.276 --> 00:09:34.903
the same denominator. They've
got to have a common

00:09:34.903 --> 00:09:38.530
denominator. The common
denominator is going to be made

00:09:38.530 --> 00:09:41.351
up of the two factors. 2 minus

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X&X +3. To write the first term
as an equivalent fraction with

00:09:46.610 --> 00:09:50.680
this denominator, we multiply
top and bottom by X plus three.

00:09:50.680 --> 00:09:56.230
So if we multiply top here by X
+3 and bottom there by X +3.

00:09:56.800 --> 00:09:58.520
Will achieve this fraction.

00:09:59.210 --> 00:10:01.613
And this fraction is equivalent
to the original 1.

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Similarly with the second term.

00:10:06.960 --> 00:10:12.630
To achieve a common denominator
of 2 minus XX +3.

00:10:13.310 --> 00:10:15.152
I need to multiply top and

00:10:15.152 --> 00:10:20.968
bottom here. By two minus X, so
B times 2 minus X and this

00:10:20.968 --> 00:10:24.808
denominator times 2 minus X and
that will give me.

00:10:25.390 --> 00:10:27.310
B2 minus X at the top.

00:10:28.420 --> 00:10:32.012
Now these two fractions
have the same denominator,

00:10:32.012 --> 00:10:36.053
we can add them together
simply by adding the

00:10:36.053 --> 00:10:39.645
numerators together, which
will give us a multiplied

00:10:39.645 --> 00:10:40.992
by X +3.

00:10:42.130 --> 00:10:47.807
Plus B multiplied by
two minus X.

00:10:48.510 --> 00:10:52.450
All divided by the common

00:10:52.450 --> 00:10:59.134
denominator. What we're saying
is that this fraction we

00:10:59.134 --> 00:11:02.699
started with is exactly the

00:11:02.699 --> 00:11:05.198
same. As this quantity here.

00:11:06.600 --> 00:11:09.484
Now the denominators
are already the same.

00:11:11.040 --> 00:11:14.784
So if this is the same as that,
and the denominators are already

00:11:14.784 --> 00:11:18.528
the same, then so too must be
the numerators, so we can equate

00:11:18.528 --> 00:11:21.984
the numerators if we equate the
numerators we can write down X

00:11:21.984 --> 00:11:26.750
equals. AX
+3.

00:11:28.280 --> 00:11:32.288
Plus B2 Minus
X.

00:11:33.710 --> 00:11:37.989
This is the equation that's
going to allow us to calculate

00:11:37.989 --> 00:11:39.156
values for A&B.

00:11:40.690 --> 00:11:44.122
Now we can find values for A&B
in one of two different ways.

00:11:44.122 --> 00:11:45.970
The 1st way that I'm going to

00:11:45.970 --> 00:11:50.400
look at. Is to substitute
specific values in for X.

00:11:51.270 --> 00:11:54.558
Remember that this quantity on
the left is supposed to be equal

00:11:54.558 --> 00:11:58.394
to this on the right for any
value of X at all. So in

00:11:58.394 --> 00:12:01.408
particular, we can choose any
values that we like. That will

00:12:01.408 --> 00:12:02.778
make all this look simpler.

00:12:03.430 --> 00:12:07.435
And what I'm going to do is I'm
going to choose X to have the

00:12:07.435 --> 00:12:09.304
value to. Why would I do that?

00:12:09.820 --> 00:12:13.744
I choose X to have the value
too, because then this second

00:12:13.744 --> 00:12:18.322
term will become zero and have 2
- 2, which is zero. Will lose

00:12:18.322 --> 00:12:20.875
this term. And we'll be able to

00:12:20.875 --> 00:12:24.740
calculate A. So by careful
choice of values for X, we can

00:12:24.740 --> 00:12:26.246
make this look a lot simpler.

00:12:27.600 --> 00:12:28.910
So with X is 2.

00:12:29.470 --> 00:12:31.276
On the left will have two.

00:12:32.110 --> 00:12:39.078
On the right will have 2 +
3, which is 5 times a.

00:12:39.520 --> 00:12:40.790
And this term will vanish.

00:12:41.620 --> 00:12:46.339
This gives me a value for a
straightaway dividing both sides

00:12:46.339 --> 00:12:50.200
by 5. I can write that a is 2/5.

00:12:52.240 --> 00:12:54.220
We need to find B.

00:12:55.660 --> 00:13:00.910
Now a sensible value that will
enable us to find B is to let X

00:13:00.910 --> 00:13:03.010
be minus three whi, is that?

00:13:03.530 --> 00:13:06.767
Well, if X is minus three,
will have minus 3 + 3, which

00:13:06.767 --> 00:13:09.257
is zero. And all of this
first term will vanish.

00:13:10.970 --> 00:13:17.410
And we'll be able to find be so
letting XP minus three will have

00:13:17.410 --> 00:13:22.930
minus three on the left zero
from this term here, and two

00:13:22.930 --> 00:13:27.990
minus minus three, which is 2 +
3, which is 55-B.

00:13:28.230 --> 00:13:35.838
Dividing both sides by 5 will
give us a value for B's

00:13:35.838 --> 00:13:38.374
minus three over 5.

00:13:39.280 --> 00:13:43.242
So now we know a value for a. We
know a value for B.

00:13:44.250 --> 00:13:47.610
And we can then proceed to
evaluate the integral by

00:13:47.610 --> 00:13:48.954
evaluating each of these

00:13:48.954 --> 00:13:56.034
separately. Let me write
this down again. We want

00:13:56.034 --> 00:14:03.564
the integral of X divided
by 2 minus XX +3.

00:14:03.600 --> 00:14:06.260
With respect to X.

00:14:06.260 --> 00:14:09.959
We expressed this algebraic
fraction in its partial fraction

00:14:09.959 --> 00:14:14.480
in its partial fractions, and we
found that a was 2/5.

00:14:15.170 --> 00:14:18.818
And be was

00:14:18.818 --> 00:14:25.024
minus 3/5. So instead
of integrating this original

00:14:25.024 --> 00:14:31.393
fraction, what we're going to do
now is integrate separately the

00:14:31.393 --> 00:14:33.130
two partial fractions.

00:14:33.250 --> 00:14:38.080
And will integrate these
separately and will do it like

00:14:38.080 --> 00:14:42.090
this. In the first integral,
we're going to take out the

00:14:42.090 --> 00:14:47.314
factor of 2/5. I will be left
with the problem of integrating

00:14:47.314 --> 00:14:50.226
one over 2 minus X with respect

00:14:50.226 --> 00:14:57.052
to X. For the second,
we're going to take out

00:14:57.052 --> 00:15:03.264
minus 3/5. And integrate
one over X +3 with

00:15:03.264 --> 00:15:05.340
respect to X.

00:15:07.710 --> 00:15:09.887
So the problem of
integrating this algebraic

00:15:09.887 --> 00:15:12.375
fraction has been split
into the problem of

00:15:12.375 --> 00:15:14.552
evaluating these two
separate integrals and both

00:15:14.552 --> 00:15:17.662
of these are simpler than
the one we started with.

00:15:18.700 --> 00:15:22.600
Let's deal with the second one
first. The second one is a

00:15:22.600 --> 00:15:26.175
situation where we've got a
function at the bottom and it's

00:15:26.175 --> 00:15:30.075
derivative at the top. Because
we've got X plus three at the

00:15:30.075 --> 00:15:34.300
bottom and the derivative of X
+3 is just one which appears at

00:15:34.300 --> 00:15:37.875
the top. So this just evaluates
to minus 3/5 the natural

00:15:37.875 --> 00:15:40.800
logarithm of the modulus of
what's at the bottom.

00:15:42.570 --> 00:15:45.468
We've got the similar situation
here, except if you

00:15:45.468 --> 00:15:48.366
differentiate the denominator,
you get minus one because of

00:15:48.366 --> 00:15:52.552
this minus X, so we'd really
like a minus one at the top.

00:15:53.620 --> 00:15:57.611
And I can adjust my numerator to
make it minus one, provided that

00:15:57.611 --> 00:16:00.681
I counteract that with putting a
minus sign outside there.

00:16:01.670 --> 00:16:06.602
So we can write all this as
minus 2/5 the natural logarithm

00:16:06.602 --> 00:16:12.767
of the modulus of 2 minus X. And
of course we need a constant of

00:16:12.767 --> 00:16:14.822
integration at the very end.

00:16:15.470 --> 00:16:20.534
So that's the result of
integrating X over 2 minus XX +3

00:16:20.534 --> 00:16:22.222
and the problems finished.

00:16:22.920 --> 00:16:28.424
What I'd like to do is just go
back a page and just show you an

00:16:28.424 --> 00:16:31.864
alternative way of calculating
values for the constants A&B in

00:16:31.864 --> 00:16:35.648
the partial fractions, and I
want us to return to this

00:16:35.648 --> 00:16:39.776
equation here that we use to
find A&B. Let me write that

00:16:39.776 --> 00:16:40.808
equation down again.

00:16:41.490 --> 00:16:47.580
X is equal to
a X +3.

00:16:47.580 --> 00:16:50.700
Plus B2 Minus

00:16:50.700 --> 00:16:58.170
X. What I'm going
to do is I'm going to

00:16:58.170 --> 00:17:01.420
start by removing the brackets.

00:17:01.650 --> 00:17:05.310
Will have a multiplied by X

00:17:05.310 --> 00:17:10.146
AX. A Times 3
which is 3A.

00:17:11.940 --> 00:17:15.258
B times two or two B.

00:17:16.130 --> 00:17:19.346
And be times minus X or
minus BX.

00:17:21.050 --> 00:17:24.768
And then what I'm going to do
is, I'm going to collect similar

00:17:24.768 --> 00:17:28.200
terms together so you see Ivan
Axe here and minus BX there.

00:17:28.770 --> 00:17:33.820
So altogether I have a minus B,
lots of X.

00:17:37.280 --> 00:17:41.739
And we've got 3A Plus
2B here.

00:17:44.620 --> 00:17:47.890
We now use this equation to
equate coefficients on both

00:17:47.890 --> 00:17:52.468
sides. What do we mean by that?
Well, what we do is we ask

00:17:52.468 --> 00:17:57.046
ourselves how many X terms do we
have on the left and match that

00:17:57.046 --> 00:18:01.624
with the number of X terms that
we have on the right. So you

00:18:01.624 --> 00:18:06.529
see, on the left hand side here,
if we look at just the ex terms,

00:18:06.529 --> 00:18:13.228
there's 1X. On the right we've
got a minus B, lots of X.

00:18:13.260 --> 00:18:18.060
So we've equated the
coefficients of X on both sides.

00:18:19.360 --> 00:18:22.948
We can also look at constant
terms on both sides. You see the

00:18:22.948 --> 00:18:24.880
three A plus 2B is a constant.

00:18:25.680 --> 00:18:30.356
There are no constant terms on
the left, so if we just look at

00:18:30.356 --> 00:18:32.360
constants, there are none on the

00:18:32.360 --> 00:18:37.360
left. And on the right there's
3A Plus 2B.

00:18:37.360 --> 00:18:40.690
And you'll see what we have.
Here are two simultaneous

00:18:40.690 --> 00:18:44.353
equations for A&B and if we
solve these equations we can

00:18:44.353 --> 00:18:48.682
find values for A&B. Let me call
that equation one and that one

00:18:48.682 --> 00:18:52.678
equation two. What I'm going to
do is I'm going to multiply

00:18:52.678 --> 00:18:57.673
equation one by two so that will
end up with two be so that we

00:18:57.673 --> 00:19:01.336
will be able to add these
together to eliminate the bees.

00:19:01.336 --> 00:19:06.331
So if I take equation one and I
multiply it by two, I'll get 2

00:19:06.331 --> 00:19:13.538
ones or two. 2A minus 2B.
Let's call that equation 3.

00:19:14.350 --> 00:19:18.028
If we add equations two and

00:19:18.028 --> 00:19:24.622
three together. We've got 0 + 2,
which is 2, three, 8 + 2 A which

00:19:24.622 --> 00:19:30.750
is 5A and two be added to minus
2B cancels out, so two is 5. In

00:19:30.750 --> 00:19:35.729
other words, A is 2 over 5,
which is the value we had

00:19:35.729 --> 00:19:37.261
earlier on for A.

00:19:37.820 --> 00:19:42.580
We can then take this value for
A and substitute it in either of

00:19:42.580 --> 00:19:46.320
these equations and obtain a
value for be. So, for example,

00:19:46.320 --> 00:19:48.360
if we substitute in the first

00:19:48.360 --> 00:19:54.790
equation. Will find that one
equals a, is 2/5 minus B.

00:19:56.100 --> 00:20:02.049
Rearranging this B is equal to
2/5 - 1.

00:20:02.790 --> 00:20:08.782
And 2/5 - 1 is minus 3/5 the
same value as we got before.

00:20:09.760 --> 00:20:13.500
So we've seen two ways of
finding the values of the

00:20:13.500 --> 00:20:17.580
constants A&B. We can substitute
specific values for X or we can

00:20:17.580 --> 00:20:18.940
equate coefficients on both

00:20:18.940 --> 00:20:23.212
sides. Often will need to use a
mix of the two methods in order

00:20:23.212 --> 00:20:24.920
to find all the constants in a

00:20:24.920 --> 00:20:31.405
given problem. Let's have
a look at

00:20:31.405 --> 00:20:34.456
a definite integral.

00:20:35.170 --> 00:20:39.031
Suppose we want to find the
integral from X is one to access

00:20:39.031 --> 00:20:46.560
2. Of three divided by
XX plus one with respect to

00:20:46.560 --> 00:20:52.800
X. As before, we examine this
integrand and ask ourselves, is

00:20:52.800 --> 00:20:55.175
this a proper or improper

00:20:55.175 --> 00:20:59.045
fraction? Well, the degree of
the denominator is too, because

00:20:59.045 --> 00:21:03.010
when we multiply this out, the
highest power of X will be 2.

00:21:03.620 --> 00:21:09.236
The degree of the numerator is
zero, with really 3X to the zero

00:21:09.236 --> 00:21:12.692
here, so this is an example of a

00:21:12.692 --> 00:21:18.092
proper fraction. On both of
these factors are linear

00:21:18.092 --> 00:21:23.960
factors. So as before, I'm going
to express the integrand.

00:21:24.580 --> 00:21:27.044
As the sum of its partial
fractions. So let's do that

00:21:27.044 --> 00:21:31.040
first of all. 3 divided by XX

00:21:31.040 --> 00:21:35.958
plus one. The appropriate form
of partial fractions.

00:21:36.930 --> 00:21:42.750
Are constant. Over the
first linear factor.

00:21:42.750 --> 00:21:47.518
Plus another constant over the
second linear factor.

00:21:49.160 --> 00:21:52.652
And our job now is to try to
find values for A&B.

00:21:53.720 --> 00:22:00.122
We do this by adding these
together as we did before,

00:22:00.122 --> 00:22:04.778
common denominator XX plus one
in both cases.

00:22:06.810 --> 00:22:10.870
To write a over X as an
equivalent fraction with

00:22:10.870 --> 00:22:14.118
this denominator will
need to multiply top and

00:22:14.118 --> 00:22:16.148
bottom by X plus one.

00:22:18.090 --> 00:22:23.927
To write B over X plus one with
this denominator will need to

00:22:23.927 --> 00:22:26.621
multiply top and bottom by X.

00:22:27.420 --> 00:22:32.840
So now we've given these two
fractions a common denominator,

00:22:32.840 --> 00:22:37.718
and we add the fractions
together by adding the

00:22:37.718 --> 00:22:43.301
numerators. I'm putting the
result over the common

00:22:43.301 --> 00:22:51.224
denominator. So 3 divided by XX
plus One is equal to all this.

00:22:53.150 --> 00:22:54.805
The denominators are already the

00:22:54.805 --> 00:23:00.050
same. So we can equate the
numerators that gives us the

00:23:00.050 --> 00:23:05.432
equation 3 equals a X plus one
plus BX. And this is the

00:23:05.432 --> 00:23:09.986
equation we can use to try to
find values for A&B.

00:23:11.260 --> 00:23:13.636
We could equate coefficients, or
we can substitute specific

00:23:13.636 --> 00:23:17.332
values for X and what I'm going
to do is I'm going to substitute

00:23:17.332 --> 00:23:21.028
the value X is not and the
reason why I'm picking X is not

00:23:21.028 --> 00:23:23.404
is because I recognize straight
away that's going to.

00:23:23.960 --> 00:23:26.552
Kill off this last term here
that'll have gone and will be

00:23:26.552 --> 00:23:28.280
able to just find a value for A.

00:23:29.280 --> 00:23:34.922
So we substitute X is not on the
left, will still have 3.

00:23:35.790 --> 00:23:41.601
And on the right we've got not
plus one which is one 1A.

00:23:41.630 --> 00:23:43.611
Be times not is not so that

00:23:43.611 --> 00:23:48.248
goes. So In other words, we've
got a value for A and a is 3.

00:23:49.640 --> 00:23:54.320
Another sensible value to
substitute is X equals minus

00:23:54.320 --> 00:23:56.565
one. Why is that a sensible

00:23:56.565 --> 00:24:00.294
value? Well, that's a sensible
value, because if we put X is

00:24:00.294 --> 00:24:03.710
minus one in minus one plus one
is zero and will lose this first

00:24:03.710 --> 00:24:05.662
term with the A in and will now

00:24:05.662 --> 00:24:09.814
be. So putting X is minus one
will have 3.

00:24:10.360 --> 00:24:14.130
This will become zero and will
have be times minus one which is

00:24:14.130 --> 00:24:19.700
minus B. So this tells us that B
is actually minus three.

00:24:20.390 --> 00:24:26.366
So now we know the value of a is
going to be 3 and B is going to

00:24:26.366 --> 00:24:30.878
be minus three. And the problem
of performing this integration

00:24:30.878 --> 00:24:34.244
can be solved by integrating
these two terms separately.

00:24:35.630 --> 00:24:41.306
Let's do that now. I'll write
these these terms down again.

00:24:41.306 --> 00:24:46.982
We're integrating three over XX
plus one with respect to X.

00:24:48.180 --> 00:24:52.310
And we've expressed already this
as its partial fractions, and

00:24:52.310 --> 00:24:56.853
found that we're integrating
three over X minus three over X,

00:24:56.853 --> 00:25:02.222
plus one with respect to X, and
this was a definite integral. It

00:25:02.222 --> 00:25:06.352
had limits on, and the limits
will one and two.

00:25:06.890 --> 00:25:10.256
So now we use partial
fractions to change this

00:25:10.256 --> 00:25:12.874
algebraic fraction into
these two simple integrals.

00:25:13.970 --> 00:25:16.088
Now, these are
straightforward to finish

00:25:16.088 --> 00:25:19.618
because the integral of
three over X is just three

00:25:19.618 --> 00:25:22.089
natural logarithm of the
modulus of X.

00:25:25.650 --> 00:25:27.408
The integral of three over X

00:25:27.408 --> 00:25:32.745
plus one. Is 3 natural logarithm
of the modulus of X plus one?

00:25:33.460 --> 00:25:35.800
And there's a minus sign
in the middle from that.

00:25:36.860 --> 00:25:40.180
This is a definite

00:25:40.180 --> 00:25:44.110
integral. So we have square
brackets and we write the limits

00:25:44.110 --> 00:25:45.660
on the right hand side.

00:25:46.350 --> 00:25:48.890
The problem is nearly
finished. All we have to do

00:25:48.890 --> 00:25:50.160
is substitute the limits in.

00:25:51.310 --> 00:25:56.590
Upper limit first when X is
2, will have three natural

00:25:56.590 --> 00:25:58.030
log of two.

00:25:59.150 --> 00:26:02.918
When X is 2 in here will have

00:26:02.918 --> 00:26:06.476
minus three. Natural
logarithm of 2 + 1, which is

00:26:06.476 --> 00:26:10.727
3. So that's what we get when
we put the upper limit in.

00:26:11.750 --> 00:26:16.524
When we put the lower limit in,
when X is one will have three

00:26:16.524 --> 00:26:17.888
natural logarithm of 1.

00:26:18.640 --> 00:26:23.398
Minus three natural logarithm of
1 + 1, which is 2. So that's

00:26:23.398 --> 00:26:28.156
what we get when we put the
lower limiting. And of course we

00:26:28.156 --> 00:26:31.450
want to find the difference of
these two quantities.

00:26:32.300 --> 00:26:36.120
Here you'll notice with three

00:26:36.120 --> 00:26:40.634
log 2. And over here
there's another three

00:26:40.634 --> 00:26:44.274
log, two with A minus and
minus, so we're adding

00:26:44.274 --> 00:26:47.550
another three log 2. So
altogether there will be

00:26:47.550 --> 00:26:48.642
6 log 2.

00:26:49.840 --> 00:26:53.220
That's minus three log 3.

00:26:53.220 --> 00:26:56.668
And the logarithm of

00:26:56.668 --> 00:27:00.540
1. Is 0 so that banishes.

00:27:01.430 --> 00:27:04.455
Now we could leave the answer
like that, although more often

00:27:04.455 --> 00:27:07.480
than not would probably use
the laws of logarithms to try

00:27:07.480 --> 00:27:10.780
to tighten this up a little
bit and write it in a

00:27:10.780 --> 00:27:13.255
different way. You should be
aware that multiplier outside,

00:27:13.255 --> 00:27:16.830
like this six, can be put
inside as a power, so we can

00:27:16.830 --> 00:27:19.580
write this as logarithm of 2
to the power 6.

00:27:21.040 --> 00:27:23.960
Subtract again a multiplier
outside can move inside as a

00:27:23.960 --> 00:27:28.048
power so we can write this as
logarithm of 3 to the power 3.

00:27:29.310 --> 00:27:32.970
And you'll also be aware from
your loss of logarithms that if

00:27:32.970 --> 00:27:36.325
we're finding the difference of
two logarithms, and we can write

00:27:36.325 --> 00:27:41.205
that as the logarithm of 2 to
the power 6 / 3 to the power 3.

00:27:41.450 --> 00:27:44.030
And that's my final answer.

00:27:45.280 --> 00:27:52.309
Let's look at another example in
which the denominator contains a

00:27:52.309 --> 00:27:57.421
repeated linear factor. Suppose
we're interested in evaluating

00:27:57.421 --> 00:28:05.089
this integral 1 divided by X
minus one all squared X Plus

00:28:05.089 --> 00:28:12.118
One, and we want to integrate
that with respect to X.

00:28:12.990 --> 00:28:14.862
So again, we have a proper

00:28:14.862 --> 00:28:19.437
fraction. And there is a linear
factor here. X plus one, another

00:28:19.437 --> 00:28:23.481
linear factor X minus one. But
this is a repeated linear factor

00:28:23.481 --> 00:28:24.829
because it occurs twice.

00:28:25.530 --> 00:28:30.795
The appropriate form of partial
fractions will be these.

00:28:31.470 --> 00:28:37.608
We want to
constant over the

00:28:37.608 --> 00:28:40.677
linear factor X

00:28:40.677 --> 00:28:47.190
minus one. We want
another constant over the linear

00:28:47.190 --> 00:28:52.690
factor repeated X minus one
squared. And finally we need

00:28:52.690 --> 00:28:58.190
another constant. See over this
linear factor X plus one.

00:28:58.190 --> 00:29:02.040
And our task is before
is to try to find values

00:29:02.040 --> 00:29:03.440
for the constants AB&C.

00:29:04.470 --> 00:29:09.042
We do that as before, by
expressing each of these over a

00:29:09.042 --> 00:29:12.471
common denominator and the
common denominator that we want

00:29:12.471 --> 00:29:15.138
is going to be X minus one.

00:29:15.700 --> 00:29:19.740
Squared X plus
one.

00:29:20.780 --> 00:29:24.560
Now to achieve a common
denominator of X minus one

00:29:24.560 --> 00:29:29.474
squared X Plus one will need to
multiply the top and bottom here

00:29:29.474 --> 00:29:35.144
by X minus 1X plus one. So we
have a X minus 1X plus one.

00:29:35.170 --> 00:29:38.455
To achieve the common
denominator in this case will

00:29:38.455 --> 00:29:42.105
need to multiply top and bottom
by X plus one.

00:29:42.620 --> 00:29:46.554
So we'll have a BX plus one.

00:29:47.120 --> 00:29:51.152
And finally, in this case, to
achieve a denominator of X minus

00:29:51.152 --> 00:29:55.856
one squared X Plus one will need
to multiply top and bottom by X

00:29:55.856 --> 00:29:56.864
minus 1 squared.

00:29:57.640 --> 00:30:04.376
Now this fraction here
is the same as

00:30:04.376 --> 00:30:06.902
this fraction here.

00:30:07.300 --> 00:30:10.479
Their denominators are already
the same, so we can equate the

00:30:10.479 --> 00:30:13.947
numerators. So if we just look
at the numerators will have one

00:30:13.947 --> 00:30:17.993
on the left is equal to the top
line here on the right hand

00:30:17.993 --> 00:30:25.160
side. AX minus
1X plus one.

00:30:25.290 --> 00:30:29.209
Plus B. X plus one.

00:30:30.290 --> 00:30:34.110
Plus C. X minus one all

00:30:34.110 --> 00:30:38.720
squared. And now we choose some
sensible values for X, so

00:30:38.720 --> 00:30:42.427
there's a lot of these terms
will drop away. For example,

00:30:42.427 --> 00:30:46.471
supposing we pick X equals 1,
what's the point of picking X

00:30:46.471 --> 00:30:50.178
equals one? Well, if we pick
axes one, this first term

00:30:50.178 --> 00:30:51.526
vanishes. We lose a.

00:30:52.220 --> 00:30:55.832
Also, if we pick X equal to 1,
the last term vanish is because

00:30:55.832 --> 00:30:59.960
we have a 1 - 1 which is zero
and will be just left with the

00:30:59.960 --> 00:31:03.930
term involving be. So by letting
XP, one will have one.

00:31:04.610 --> 00:31:07.634
On the left is equal to 0.

00:31:08.200 --> 00:31:12.070
One and one here is 22B.

00:31:12.070 --> 00:31:19.090
And the last term vanishes. In
other words, B is equal to 1/2.

00:31:20.380 --> 00:31:24.982
What's another sensible value to
pick for X? Well, if we let X

00:31:24.982 --> 00:31:26.044
equal minus one.

00:31:26.780 --> 00:31:30.641
X being minus one will mean that
this term vanish is minus one

00:31:30.641 --> 00:31:31.829
plus One is 0.

00:31:32.520 --> 00:31:36.953
This term will vanish minus one,
plus one is zero and will be

00:31:36.953 --> 00:31:41.386
able to find see. So I'm going
to let X be minus one.

00:31:41.620 --> 00:31:44.707
Will still have the one on the

00:31:44.707 --> 00:31:47.890
left. When X is minus one, this

00:31:47.890 --> 00:31:50.110
goes. This goes.

00:31:50.680 --> 00:31:55.375
And on the right hand side this
term here will have minus 1 - 1

00:31:55.375 --> 00:31:59.757
is minus two we square it will
get plus four, so will get plus

00:31:59.757 --> 00:32:04.999
4C. In other words, see is going
to be 1/4.

00:32:06.070 --> 00:32:11.110
So we've got be. We've got C and
now we need to find a value for

00:32:11.110 --> 00:32:15.205
a. Now we can substitute any
other value we like in here, so

00:32:15.205 --> 00:32:18.985
I'm actually going to pick X
equals 0. It's a nice simple

00:32:18.985 --> 00:32:23.080
value. Effects is zero, will
have one on the left if X is

00:32:23.080 --> 00:32:26.230
zero there and there will be
left with minus one.

00:32:26.790 --> 00:32:32.404
Minus 1 * 1 is minus 1 - 1
times a is minus a.

00:32:33.100 --> 00:32:38.840
Thanks is 0 here. Will just left
with B Times one which is be.

00:32:38.850 --> 00:32:44.422
And if X is 0 here will have
minus one squared, which is plus

00:32:44.422 --> 00:32:50.392
one plus One Times C Plus C. Now
we already know values for B and

00:32:50.392 --> 00:32:55.964
for C, so we substitute these in
will have one is equal to minus

00:32:55.964 --> 00:33:00.342
A plus B which is 1/2 plus C
which is 1/4.

00:33:01.490 --> 00:33:06.150
So rearranging this will have
that a is equal to.

00:33:06.730 --> 00:33:12.260
Well, with a half and a quarter,
that's 3/4 and the one from this

00:33:12.260 --> 00:33:17.790
side over the other side is
minus one. 3/4 - 1 is going to

00:33:17.790 --> 00:33:18.975
be minus 1/4.

00:33:18.980 --> 00:33:25.972
So there we have our values for
a for B and for C, so a being

00:33:25.972 --> 00:33:29.031
minus 1/4 will go back in there.

00:33:29.330 --> 00:33:34.062
Be being a half will go back in
here and see being a quarter

00:33:34.062 --> 00:33:38.118
will go back in there and then
we'll have three separate pieces

00:33:38.118 --> 00:33:41.498
of integration to do in order to
complete the problem.

00:33:42.460 --> 00:33:46.468
Let me write these all down
again. I'm going to write the

00:33:46.468 --> 00:33:49.474
integral out and write these
three separate ones down.

00:33:49.490 --> 00:33:56.826
So we had that the integral of
one over X minus one squared X

00:33:56.826 --> 00:34:03.114
Plus one, all integrated with
respect to X, is going to equal.

00:34:03.860 --> 00:34:09.940
Minus 1/4 the integral of one
over X minus one.

00:34:10.610 --> 00:34:17.918
They'll be plus 1/2 integral of
one over X minus 1 squared.

00:34:18.480 --> 00:34:24.384
And they'll
be 1/4

00:34:24.384 --> 00:34:27.336
over X

00:34:27.336 --> 00:34:33.425
plus one.
And we want

00:34:33.425 --> 00:34:36.500
to DX in

00:34:36.500 --> 00:34:42.033
every term. So we've
used partial fractions to split

00:34:42.033 --> 00:34:45.823
this integrand into three
separate terms, and will try and

00:34:45.823 --> 00:34:49.613
finish this off. Now. The first
integral straightforward when we

00:34:49.613 --> 00:34:54.161
integrate one over X minus one
will end up with just the

00:34:54.161 --> 00:34:57.572
natural logarithm of the
denominator, so we'll end up

00:34:57.572 --> 00:35:00.983
with minus 1/4 natural
logarithm. The modulus of X

00:35:00.983 --> 00:35:06.208
minus one. To integrate this
term, we're integrating one over

00:35:06.208 --> 00:35:10.618
X minus one squared. Let me just
do that separately.

00:35:10.770 --> 00:35:16.362
To integrate one over X minus
one all squared, we make a

00:35:16.362 --> 00:35:20.090
substitution and let you equals
X minus one.

00:35:20.760 --> 00:35:25.282
Do you then will equal du DX,
which is just one times DX, so

00:35:25.282 --> 00:35:27.866
do you will be the same as DX?

00:35:28.450 --> 00:35:32.239
The integral will become the
integral of one over.

00:35:32.980 --> 00:35:35.676
X minus one squared
will be you squared.

00:35:36.820 --> 00:35:38.440
And DX is D.

00:35:39.210 --> 00:35:42.666
Now, this is straightforward to
finish because this is

00:35:42.666 --> 00:35:44.970
integrating you to the minus 2.

00:35:45.630 --> 00:35:49.040
Increase the power by one
becomes you to the minus 1

00:35:49.040 --> 00:35:50.590
divided by the new power.

00:35:51.090 --> 00:35:52.806
And add a constant of
integration.

00:35:54.010 --> 00:35:59.197
So when we do this, integration
will end up with minus one over

00:35:59.197 --> 00:36:01.480
you. Which is minus one over.

00:36:02.250 --> 00:36:04.008
X minus one.

00:36:04.660 --> 00:36:10.861
And we can put that back into
here now. So the integral half

00:36:10.861 --> 00:36:16.108
integral of one over X minus one
squared will be 1/2.

00:36:16.210 --> 00:36:19.876
All that we've got down here,
which is minus one over X minus

00:36:19.876 --> 00:36:24.960
one. The constant of integration
will add another very end. This

00:36:24.960 --> 00:36:27.318
integral is going to be 1/4.

00:36:27.890 --> 00:36:31.894
The natural logarithm of the
modulus of X Plus One and then a

00:36:31.894 --> 00:36:33.742
single constant for all of that.

00:36:35.110 --> 00:36:39.491
Let me just tidy up this little
a little bit. The two logarithm

00:36:39.491 --> 00:36:43.198
terms can be combined. We've got
a quarter of this logarithm.

00:36:43.198 --> 00:36:46.905
Subtract 1/4 of this logarithm,
so together we've got a quarter

00:36:46.905 --> 00:36:50.612
the logarithm. If we use the
laws of logarithms, we've gotta

00:36:50.612 --> 00:36:54.656
log subtracted log. So we want
the first term X Plus one

00:36:54.656 --> 00:36:56.678
divided by the second term X

00:36:56.678 --> 00:37:03.280
minus one. And then this term
can be written as minus 1/2.

00:37:03.280 --> 00:37:06.060
One over X minus one.

00:37:06.060 --> 00:37:09.954
Is a constant of integration
at the end, and that's the

00:37:09.954 --> 00:37:10.662
integration complete.

00:37:11.960 --> 00:37:18.440
Now let's look at an example in
which we have an improper

00:37:18.440 --> 00:37:21.270
fraction. Suppose we have this

00:37:21.270 --> 00:37:27.474
integral. The
degree of

00:37:27.474 --> 00:37:30.730
the numerator

00:37:30.730 --> 00:37:36.332
is 3.
The degree of the

00:37:36.332 --> 00:37:37.556
denominator is 2.

00:37:38.610 --> 00:37:41.620
3 being greater than two
means that this is an

00:37:41.620 --> 00:37:42.222
improper fraction.

00:37:43.330 --> 00:37:46.386
Improper fractions requires
special treatment and the first

00:37:46.386 --> 00:37:49.824
thing we do is division
polynomial division. Now, if

00:37:49.824 --> 00:37:53.644
you're not happy with Long
Division of polynomials, then I

00:37:53.644 --> 00:37:57.464
would suggest that you look
again at the video called

00:37:57.464 --> 00:38:00.868
polynomial division. When
examples like this are done very

00:38:00.868 --> 00:38:05.124
thoroughly, but what we want to
do is see how many times X minus

00:38:05.124 --> 00:38:06.948
X squared minus four will divide

00:38:06.948 --> 00:38:13.470
into X cubed. The way we do
this polynomial division is we

00:38:13.470 --> 00:38:19.630
say how many times does X
squared divided into X cubed.

00:38:20.330 --> 00:38:24.878
That's like asking How many
times X squared will go into X

00:38:24.878 --> 00:38:29.047
cubed, and clearly when X
squared is divided into X cubed,

00:38:29.047 --> 00:38:30.942
the answer is just X.

00:38:31.920 --> 00:38:35.652
So X squared goes into X
cubed X times and we write

00:38:35.652 --> 00:38:36.896
the solution down there.

00:38:38.130 --> 00:38:42.186
We take what we have just
written down and multiply it by

00:38:42.186 --> 00:38:45.566
everything here. So X times X
squared is X cubed.

00:38:46.130 --> 00:38:49.588
X times minus four is minus 4X.

00:38:50.370 --> 00:38:51.878
And then we subtract.

00:38:52.690 --> 00:38:55.469
X cubed minus X cubed vanish is.

00:38:56.040 --> 00:39:00.385
With no access here, and we're
subtracting minus 4X, which is

00:39:00.385 --> 00:39:01.570
like adding 4X.

00:39:03.110 --> 00:39:08.141
This means that when we divide X
squared minus four into X cubed,

00:39:08.141 --> 00:39:13.559
we get a whole part X and a
remainder 4X. In other words, X

00:39:13.559 --> 00:39:18.203
cubed divided by X squared minus
four can be written as X.

00:39:19.360 --> 00:39:25.096
Plus 4X divided by X
squared minus 4.

00:39:25.270 --> 00:39:30.242
So in order to tackle this
integration, we've done the long

00:39:30.242 --> 00:39:34.762
division and we're left with two
separate integrals to workout.

00:39:35.640 --> 00:39:37.980
Now this one is going to be
straightforward. Clearly you're

00:39:37.980 --> 00:39:41.022
just integrating X, it's going
to be X squared over 2. This is

00:39:41.022 --> 00:39:43.596
a bit more problematic. Let's
have a look at this again.

00:39:44.420 --> 00:39:48.083
This is now a proper fraction
where the degree of the

00:39:48.083 --> 00:39:51.413
numerator is one with the next
to the one here.

00:39:52.850 --> 00:39:54.512
The degree of the denominator is

00:39:54.512 --> 00:39:56.810
2. So it's a proper fraction.

00:39:58.130 --> 00:40:00.990
It looks as though we've got a
quadratic factor in the

00:40:00.990 --> 00:40:05.690
denominator. But in fact, X
squared minus four will

00:40:05.690 --> 00:40:10.420
factorize. It's in fact, the
difference of two squares. So we

00:40:10.420 --> 00:40:16.440
can write 4X over X squared
minus four as 4X over X minus 2X

00:40:16.440 --> 00:40:20.740
plus two. So the quadratic will
actually factorize and you'll

00:40:20.740 --> 00:40:23.320
see. Now we've got two linear

00:40:23.320 --> 00:40:28.483
factors. We can express this in
partial fractions. Let's do

00:40:28.483 --> 00:40:35.021
that. We've got 4X over X, minus
two X +2, and because each of

00:40:35.021 --> 00:40:40.158
these are linear factors, the
appropriate form is going to be

00:40:40.158 --> 00:40:42.493
a over X minus 2.

00:40:43.160 --> 00:40:46.980
Plus B over X +2.

00:40:49.520 --> 00:40:54.990
We add these together using a
common dumped common denominator

00:40:54.990 --> 00:41:02.648
X minus two X +2 and will
get a X +2 plus BX minus

00:41:02.648 --> 00:41:09.210
2. All over the common
denominator, X minus two X +2.

00:41:12.310 --> 00:41:18.988
Now the left hand side and the
right hand side of the same. The

00:41:18.988 --> 00:41:23.758
denominators are already the
same, so the numerators must be

00:41:23.758 --> 00:41:27.574
the same. 4X must equal a X +2

00:41:27.574 --> 00:41:29.790
plus B. 6 - 2.

00:41:30.850 --> 00:41:35.708
This is the equation that we can
use to find the values for A&B.

00:41:37.550 --> 00:41:39.458
Let me write that down again.

00:41:39.990 --> 00:41:45.758
We've got 4X is
a X +2.

00:41:46.440 --> 00:41:51.020
Plus BX minus
2.

00:41:52.630 --> 00:41:56.844
What's a sensible value to put
in for X? Well, if we choose, X

00:41:56.844 --> 00:41:58.951
is 2, will lose the second term.

00:41:59.620 --> 00:42:02.728
So X is 2 is a good
value to put in here.

00:42:03.940 --> 00:42:09.100
Faxes 2 on the left hand side
will have 4 * 2, which is 8.

00:42:10.070 --> 00:42:15.410
If X is 2, will have 2 +
2, which is 44A.

00:42:15.410 --> 00:42:17.940
And this term will vanish.

00:42:19.000 --> 00:42:22.200
So 8 equals 4A.

00:42:22.380 --> 00:42:27.340
A will equal 2 and that's the
value for A.

00:42:29.200 --> 00:42:32.684
What's another sensible value to
put in for X? Well, if X is

00:42:32.684 --> 00:42:36.436
minus two, will have minus 2 +
2, which is zero. Will lose this

00:42:36.436 --> 00:42:44.200
term. So if X is
minus two, will have minus 8

00:42:44.200 --> 00:42:46.180
on the left.

00:42:46.190 --> 00:42:47.710
This first term will vanish.

00:42:48.490 --> 00:42:54.490
An ex being minus two here will
have minus 2 - 2 is minus four

00:42:54.490 --> 00:42:55.690
so minus 4B.

00:42:55.690 --> 00:43:01.114
So be must also be
equal to two.

00:43:01.120 --> 00:43:06.160
So now we've got values for A
and for be both equal to two.

00:43:06.160 --> 00:43:08.680
Let's take us back and see what

00:43:08.680 --> 00:43:11.110
that means. It means that when

00:43:11.110 --> 00:43:14.736
we express. This quantity
4X over X squared minus

00:43:14.736 --> 00:43:17.868
four in partial fractions
like this, the values of

00:43:17.868 --> 00:43:19.956
A&B are both equal to two.

00:43:21.140 --> 00:43:26.740
So the integral were working out
is the integral of X +2 over X

00:43:26.740 --> 00:43:29.540
minus 2 + 2 over X +2.

00:43:29.550 --> 00:43:35.694
So there we
are. We've now

00:43:35.694 --> 00:43:41.838
got three separate
integrals to evaluate,

00:43:41.838 --> 00:43:47.982
and it's straightforward
to finish this

00:43:47.982 --> 00:43:54.540
off. The integral of X
is X squared divided by two.

00:43:55.650 --> 00:44:02.475
The integral of two over X minus
2 equals 2 natural logarithm of

00:44:02.475 --> 00:44:05.625
the modulus of X minus 2.

00:44:06.430 --> 00:44:13.934
And the integral of two over X
+2 is 2 natural logarithm of X

00:44:13.934 --> 00:44:16.614
+2 plus a constant of

00:44:16.614 --> 00:44:20.947
integration. If we wanted to do,
we could use the laws of

00:44:20.947 --> 00:44:23.617
logarithms to combine these log
rhythmic terms here, but I'll

00:44:23.617 --> 00:44:24.952
leave the answer like that.