0:00:00.900,0:00:04.540
In this video, we're going to[br]have a look at how we can

0:00:04.540,0:00:06.721
integrate algebraic[br]fractions. The sorts of

0:00:06.721,0:00:09.214
fractions that we're going to[br]integrate and like these

0:00:09.214,0:00:09.491
here.

0:00:10.590,0:00:13.990
Now, superficially, they will[br]look very similar, but there are

0:00:13.990,0:00:17.730
important differences which I'd[br]like to point out when you come

0:00:17.730,0:00:20.790
to tackle a problem of[br]integrating a fraction like

0:00:20.790,0:00:24.190
this, it's important that you[br]can look for certain features,

0:00:24.190,0:00:25.890
for example, in this first

0:00:25.890,0:00:30.610
example. In the denominator we[br]have what we call 2 linear

0:00:30.610,0:00:33.976
factors. These two linear[br]factors are two different linear

0:00:33.976,0:00:38.090
factors. When I say linear[br]factors, I mean there's no X

0:00:38.090,0:00:42.578
squared, no ex cubes, nothing[br]like that in it. These are just

0:00:42.578,0:00:47.814
of the form a X Plus B linear[br]factors. A constant times X plus

0:00:47.814,0:00:50.806
another constant. So with two[br]linear factors here.

0:00:51.870,0:00:55.965
This example is also got linear[br]factors in clearly X Plus One is

0:00:55.965,0:01:00.720
a linear factor. X minus one is[br]a linear factor, but the fact

0:01:00.720,0:01:04.440
that I've got an X minus one[br]squared means that we've really

0:01:04.440,0:01:08.470
got X minus one times X minus 1[br]two linear factors within here.

0:01:08.470,0:01:11.880
So we call this an example of a[br]repeated linear factor.

0:01:13.150,0:01:18.600
Linear factors, repeated linear[br]factors. Over here we've got a

0:01:18.600,0:01:23.505
quadratic. Now this particular[br]quadratic will not factorize and

0:01:23.505,0:01:28.955
because it won't factorize, we[br]call it an irreducible quadratic

0:01:28.955,0:01:33.700
factor. This final example here[br]has also got a quadratic factor

0:01:33.700,0:01:37.330
in the denominator, but unlike[br]the previous one, this one will

0:01:37.330,0:01:40.630
in fact factorize because X[br]squared minus four is actually

0:01:40.630,0:01:44.920
the difference of two squares[br]and we can write this as X minus

0:01:44.920,0:01:48.880
2X plus two. So whilst it might[br]have originally looked like a

0:01:48.880,0:01:52.510
quadratic factor, it was in fact[br]to linear factors, so that's

0:01:52.510,0:01:56.140
what those are one of the[br]important things we should be

0:01:56.140,0:01:59.440
looking for when we come to[br]integrate quantities like these.

0:01:59.440,0:02:00.760
Weather we've got linear

0:02:00.760,0:02:03.535
factors. Repeated linear[br]factors, irreducible quadratic

0:02:03.535,0:02:06.902
factors, or quadratic factors[br]that will factorize.

0:02:07.440,0:02:10.560
Something else which is[br]important as well, is to examine

0:02:10.560,0:02:14.304
the degree of the numerator and[br]the degree of the denominator in

0:02:14.304,0:02:17.112
each of these fractions.[br]Remember, the degree is the

0:02:17.112,0:02:20.544
highest power, so for example in[br]the denominator of this example

0:02:20.544,0:02:24.288
here, if we multiplied it all[br]out, we actually get the highest

0:02:24.288,0:02:27.720
power as three, because when we[br]multiply the first terms out

0:02:27.720,0:02:31.776
will get an X squared, and when[br]we multiply it with the SEC

0:02:31.776,0:02:36.144
bracket, X Plus, one will end up[br]with an X cubed. So the degree

0:02:36.144,0:02:38.016
of the denominator there is 3.

0:02:38.670,0:02:42.954
The degree of the numerator is 0[br]because we can think of this as

0:02:42.954,0:02:44.484
One X to the 0.

0:02:46.040,0:02:50.345
In the first case, we've got an[br]X to the one here, so the degree

0:02:50.345,0:02:53.789
of the numerator there is one,[br]and if we multiply the brackets

0:02:53.789,0:02:57.233
out, the degree of the of the[br]denominator will be two. Will

0:02:57.233,0:03:00.964
get a quadratic term in here. In[br]this case, the degree of the

0:03:00.964,0:03:05.262
numerator is 0. And in this[br]case, the degree of the

0:03:05.262,0:03:09.630
denominator is to the highest[br]power is too. So in all of

0:03:09.630,0:03:13.634
these cases, the degree of the[br]numerator is less than the

0:03:13.634,0:03:17.274
degree of the denominator, and[br]we call fractions like these

0:03:17.274,0:03:18.002
proper fractions.

0:03:20.110,0:03:24.114
On the other hand, if we look at[br]this final example, the degree

0:03:24.114,0:03:27.810
of the numerator is 3, whereas[br]the degree of the denominator is

0:03:27.810,0:03:31.814
too. So in this case, the degree[br]of the numerator is greater than

0:03:31.814,0:03:35.202
the degree of the denominator,[br]and this is what's called an

0:03:35.202,0:03:38.670
improper fraction. Now when we[br]start to integrate quantities

0:03:38.670,0:03:41.970
like this will need to examine[br]whether we're dealing with

0:03:41.970,0:03:45.270
proper fractions or improper[br]fractions, and then, as I said

0:03:45.270,0:03:49.230
before, will need to look at all[br]the factors in the denominator.

0:03:49.800,0:03:53.485
Will also need to call appan[br]techniques in the theory of

0:03:53.485,0:03:56.835
partial fractions. There is a[br]video on partial fractions and

0:03:56.835,0:03:59.515
you may wish to refer to that if

0:03:59.515,0:04:04.090
necessary. If you have a linear[br]factor in the denominator, this

0:04:04.090,0:04:08.254
will lead to a partial fraction[br]of this form. A constant over

0:04:08.254,0:04:09.295
the linear factor.

0:04:10.100,0:04:13.740
If you have a repeated linear[br]factor in the denominator,

0:04:13.740,0:04:16.652
you'll need two partial[br]fractions. A constant over

0:04:16.652,0:04:19.928
the factor and a constant[br]over the factor squared.

0:04:21.840,0:04:25.460
Finally, if you have a quadratic[br]factor which is irreducible,

0:04:25.460,0:04:29.804
you'll need to write a partial[br]fraction of the form a constant

0:04:29.804,0:04:33.062
times X plus another constant[br]over the irreducible quadratic

0:04:33.062,0:04:36.682
factor, so will certainly be[br]calling upon the techniques of

0:04:36.682,0:04:41.616
partial fractions. Will also[br]need to call appan. Lots of

0:04:41.616,0:04:42.987
techniques and integration.

0:04:42.990,0:04:46.554
I'm just going to mention just[br]two or three here, which will

0:04:46.554,0:04:49.821
need to use as we proceed[br]through the examples of 1st.

0:04:49.821,0:04:53.088
Crucial result is the standard[br]result, which says that if you

0:04:53.088,0:04:56.058
have an integral consisting of a[br]function in the denominator.

0:04:56.770,0:05:00.810
And it's derivative in the[br]numerator. Then the result is

0:05:00.810,0:05:05.254
the logarithm of the modulus of[br]the function in the denominator.

0:05:05.254,0:05:10.910
So for example, if I ask you to[br]integrate one over X plus one

0:05:10.910,0:05:12.526
with respect to X.

0:05:13.320,0:05:16.818
Then clearly the function in the[br]denominator is X plus one.

0:05:17.390,0:05:20.668
And its derivative is one which[br]appears in the numerator. So

0:05:20.668,0:05:22.456
we've an example of this form.

0:05:23.640,0:05:27.372
So the resulting integral is the[br]logarithm of the modulus of the

0:05:27.372,0:05:30.482
function that was in the[br]denominator, which is X plus

0:05:30.482,0:05:34.321
one. Plus a constant of[br]integration, so we will need

0:05:34.321,0:05:37.830
that result very frequently in[br]the examples which are going to

0:05:37.830,0:05:41.339
follow will also need some[br]standard results and one of the

0:05:41.339,0:05:44.848
standard results I will call[br]appan is this one. The integral

0:05:44.848,0:05:48.995
of one over a squared plus X[br]squared is one over a inverse

0:05:48.995,0:05:51.228
tan of X over a plus C.

0:05:52.400,0:05:55.821
Results like this can be found[br]in tables of standard integrals.

0:05:57.120,0:06:00.850
Finally, we need to integrate[br]quantities like this and you'll

0:06:00.850,0:06:04.580
need to do this probably using[br]integration by substitution. An

0:06:04.580,0:06:08.683
integral like this can be worked[br]out by making the substitution

0:06:08.683,0:06:10.548
you equals X minus one.

0:06:11.260,0:06:16.604
So that the differential du[br]is du DX.

0:06:16.610,0:06:22.790
DX, which in this case is du[br]DX, will be just one.

0:06:22.850,0:06:27.470
So do you is DX and that's[br]integral. Then will become

0:06:27.470,0:06:30.830
the integral of one over[br]you, squared du.

0:06:32.080,0:06:36.588
One over you squared is the same[br]as the integral of you to the

0:06:36.588,0:06:40.130
minus 2D U, which you can solve[br]by integrating increasing the

0:06:40.130,0:06:44.638
power BI want to give you you to[br]the minus one over minus one.

0:06:45.290,0:06:46.770
Plus a constant of integration.

0:06:48.180,0:06:51.912
This can be finished off by[br]changing the you back to the

0:06:51.912,0:06:55.644
original variable X minus one[br]and that will give us X minus

0:06:55.644,0:06:57.821
one to the minus one over minus

0:06:57.821,0:07:02.868
one plus C. Which is the same as[br]minus one over X minus one plus

0:07:02.868,0:07:04.891
C, which is the results I have

0:07:04.891,0:07:09.384
here. So what I'm saying is that[br]throughout the rest of this unit

0:07:09.384,0:07:12.708
will need to call Appan lots of[br]different techniques to be able

0:07:12.708,0:07:13.816
to perform the integrals.

0:07:14.540,0:07:16.020
As we shall see.

0:07:16.590,0:07:19.902
Let's look at the

0:07:19.902,0:07:26.492
first example. Suppose we[br]want to integrate this algebraic

0:07:26.492,0:07:29.772
fraction. 6 / 2

0:07:29.772,0:07:33.245
minus X. X

0:07:33.245,0:07:38.140
+3[br]DX

0:07:39.300,0:07:42.990
The first thing we do is we look[br]at the object we've got and try

0:07:42.990,0:07:45.696
to ask ourselves, are we dealing[br]with a proper or improper

0:07:45.696,0:07:48.156
fraction and what are the[br]factors in the denominator like?

0:07:49.200,0:07:53.208
Well, if we multiply the power,[br]the brackets at the bottom will

0:07:53.208,0:07:57.884
find that the highest power of X[br]is X squared, so the degree of

0:07:57.884,0:07:59.220
the denominator is 2.

0:08:00.430,0:08:04.318
The highest power in the[br]numerator is one. This is an X

0:08:04.318,0:08:08.530
to the power one, so the degree[br]of the numerator is one because

0:08:08.530,0:08:12.418
the degree of the numerator is[br]less than the degree of the

0:08:12.418,0:08:16.460
denominator. This is an example[br]of a proper fraction.

0:08:17.040,0:08:22.745
Both of these factors[br]in the denominator.

0:08:23.380,0:08:24.658
Are linear factors.

0:08:25.250,0:08:31.800
So we're dealing with a proper[br]fraction with linear factors.

0:08:31.820,0:08:37.787
The way we proceed is to take[br]this fraction and express it in

0:08:37.787,0:08:41.918
partial fractions. So I'll start[br]with the fraction again.

0:08:42.770,0:08:49.043
And express it in the[br]appropriate form of partial

0:08:49.043,0:08:51.831
fractions. Now because it's

0:08:51.831,0:08:56.206
proper. And because we've got[br]linear factors, the appropriate

0:08:56.206,0:09:00.573
form is to have a constant over[br]the first linear factor.

0:09:01.560,0:09:06.376
Plus another constant over the[br]second linear factor.

0:09:07.580,0:09:11.529
Our task now is to find values[br]for the constants A&B.

0:09:12.220,0:09:15.980
Once we've done that, will[br]be able to evaluate this

0:09:15.980,0:09:18.236
integral by evaluating these[br]two separately.

0:09:19.480,0:09:24.145
So to find A and be the first[br]thing we do is we add these

0:09:24.145,0:09:25.389
two fractions together again.

0:09:26.440,0:09:31.276
Remember that to add 2 fractions[br]together, we've got to give them

0:09:31.276,0:09:34.903
the same denominator. They've[br]got to have a common

0:09:34.903,0:09:38.530
denominator. The common[br]denominator is going to be made

0:09:38.530,0:09:41.351
up of the two factors. 2 minus

0:09:41.351,0:09:46.610
X&X +3. To write the first term[br]as an equivalent fraction with

0:09:46.610,0:09:50.680
this denominator, we multiply[br]top and bottom by X plus three.

0:09:50.680,0:09:56.230
So if we multiply top here by X[br]+3 and bottom there by X +3.

0:09:56.800,0:09:58.520
Will achieve this fraction.

0:09:59.210,0:10:01.613
And this fraction is equivalent[br]to the original 1.

0:10:03.550,0:10:06.370
Similarly with the second term.

0:10:06.960,0:10:12.630
To achieve a common denominator[br]of 2 minus XX +3.

0:10:13.310,0:10:15.152
I need to multiply top and

0:10:15.152,0:10:20.968
bottom here. By two minus X, so[br]B times 2 minus X and this

0:10:20.968,0:10:24.808
denominator times 2 minus X and[br]that will give me.

0:10:25.390,0:10:27.310
B2 minus X at the top.

0:10:28.420,0:10:32.012
Now these two fractions[br]have the same denominator,

0:10:32.012,0:10:36.053
we can add them together[br]simply by adding the

0:10:36.053,0:10:39.645
numerators together, which[br]will give us a multiplied

0:10:39.645,0:10:40.992
by X +3.

0:10:42.130,0:10:47.807
Plus B multiplied by[br]two minus X.

0:10:48.510,0:10:52.450
All divided by the common

0:10:52.450,0:10:59.134
denominator. What we're saying[br]is that this fraction we

0:10:59.134,0:11:02.699
started with is exactly the

0:11:02.699,0:11:05.198
same. As this quantity here.

0:11:06.600,0:11:09.484
Now the denominators[br]are already the same.

0:11:11.040,0:11:14.784
So if this is the same as that,[br]and the denominators are already

0:11:14.784,0:11:18.528
the same, then so too must be[br]the numerators, so we can equate

0:11:18.528,0:11:21.984
the numerators if we equate the[br]numerators we can write down X

0:11:21.984,0:11:26.750
equals. AX[br]+3.

0:11:28.280,0:11:32.288
Plus B2 Minus[br]X.

0:11:33.710,0:11:37.989
This is the equation that's[br]going to allow us to calculate

0:11:37.989,0:11:39.156
values for A&B.

0:11:40.690,0:11:44.122
Now we can find values for A&B[br]in one of two different ways.

0:11:44.122,0:11:45.970
The 1st way that I'm going to

0:11:45.970,0:11:50.400
look at. Is to substitute[br]specific values in for X.

0:11:51.270,0:11:54.558
Remember that this quantity on[br]the left is supposed to be equal

0:11:54.558,0:11:58.394
to this on the right for any[br]value of X at all. So in

0:11:58.394,0:12:01.408
particular, we can choose any[br]values that we like. That will

0:12:01.408,0:12:02.778
make all this look simpler.

0:12:03.430,0:12:07.435
And what I'm going to do is I'm[br]going to choose X to have the

0:12:07.435,0:12:09.304
value to. Why would I do that?

0:12:09.820,0:12:13.744
I choose X to have the value[br]too, because then this second

0:12:13.744,0:12:18.322
term will become zero and have 2[br]- 2, which is zero. Will lose

0:12:18.322,0:12:20.875
this term. And we'll be able to

0:12:20.875,0:12:24.740
calculate A. So by careful[br]choice of values for X, we can

0:12:24.740,0:12:26.246
make this look a lot simpler.

0:12:27.600,0:12:28.910
So with X is 2.

0:12:29.470,0:12:31.276
On the left will have two.

0:12:32.110,0:12:39.078
On the right will have 2 +[br]3, which is 5 times a.

0:12:39.520,0:12:40.790
And this term will vanish.

0:12:41.620,0:12:46.339
This gives me a value for a[br]straightaway dividing both sides

0:12:46.339,0:12:50.200
by 5. I can write that a is 2/5.

0:12:52.240,0:12:54.220
We need to find B.

0:12:55.660,0:13:00.910
Now a sensible value that will[br]enable us to find B is to let X

0:13:00.910,0:13:03.010
be minus three whi, is that?

0:13:03.530,0:13:06.767
Well, if X is minus three,[br]will have minus 3 + 3, which

0:13:06.767,0:13:09.257
is zero. And all of this[br]first term will vanish.

0:13:10.970,0:13:17.410
And we'll be able to find be so[br]letting XP minus three will have

0:13:17.410,0:13:22.930
minus three on the left zero[br]from this term here, and two

0:13:22.930,0:13:27.990
minus minus three, which is 2 +[br]3, which is 55-B.

0:13:28.230,0:13:35.838
Dividing both sides by 5 will[br]give us a value for B's

0:13:35.838,0:13:38.374
minus three over 5.

0:13:39.280,0:13:43.242
So now we know a value for a. We[br]know a value for B.

0:13:44.250,0:13:47.610
And we can then proceed to[br]evaluate the integral by

0:13:47.610,0:13:48.954
evaluating each of these

0:13:48.954,0:13:56.034
separately. Let me write[br]this down again. We want

0:13:56.034,0:14:03.564
the integral of X divided[br]by 2 minus XX +3.

0:14:03.600,0:14:06.260
With respect to X.

0:14:06.260,0:14:09.959
We expressed this algebraic[br]fraction in its partial fraction

0:14:09.959,0:14:14.480
in its partial fractions, and we[br]found that a was 2/5.

0:14:15.170,0:14:18.818
And be was

0:14:18.818,0:14:25.024
minus 3/5. So instead[br]of integrating this original

0:14:25.024,0:14:31.393
fraction, what we're going to do[br]now is integrate separately the

0:14:31.393,0:14:33.130
two partial fractions.

0:14:33.250,0:14:38.080
And will integrate these[br]separately and will do it like

0:14:38.080,0:14:42.090
this. In the first integral,[br]we're going to take out the

0:14:42.090,0:14:47.314
factor of 2/5. I will be left[br]with the problem of integrating

0:14:47.314,0:14:50.226
one over 2 minus X with respect

0:14:50.226,0:14:57.052
to X. For the second,[br]we're going to take out

0:14:57.052,0:15:03.264
minus 3/5. And integrate[br]one over X +3 with

0:15:03.264,0:15:05.340
respect to X.

0:15:07.710,0:15:09.887
So the problem of[br]integrating this algebraic

0:15:09.887,0:15:12.375
fraction has been split[br]into the problem of

0:15:12.375,0:15:14.552
evaluating these two[br]separate integrals and both

0:15:14.552,0:15:17.662
of these are simpler than[br]the one we started with.

0:15:18.700,0:15:22.600
Let's deal with the second one[br]first. The second one is a

0:15:22.600,0:15:26.175
situation where we've got a[br]function at the bottom and it's

0:15:26.175,0:15:30.075
derivative at the top. Because[br]we've got X plus three at the

0:15:30.075,0:15:34.300
bottom and the derivative of X[br]+3 is just one which appears at

0:15:34.300,0:15:37.875
the top. So this just evaluates[br]to minus 3/5 the natural

0:15:37.875,0:15:40.800
logarithm of the modulus of[br]what's at the bottom.

0:15:42.570,0:15:45.468
We've got the similar situation[br]here, except if you

0:15:45.468,0:15:48.366
differentiate the denominator,[br]you get minus one because of

0:15:48.366,0:15:52.552
this minus X, so we'd really[br]like a minus one at the top.

0:15:53.620,0:15:57.611
And I can adjust my numerator to[br]make it minus one, provided that

0:15:57.611,0:16:00.681
I counteract that with putting a[br]minus sign outside there.

0:16:01.670,0:16:06.602
So we can write all this as[br]minus 2/5 the natural logarithm

0:16:06.602,0:16:12.767
of the modulus of 2 minus X. And[br]of course we need a constant of

0:16:12.767,0:16:14.822
integration at the very end.

0:16:15.470,0:16:20.534
So that's the result of[br]integrating X over 2 minus XX +3

0:16:20.534,0:16:22.222
and the problems finished.

0:16:22.920,0:16:28.424
What I'd like to do is just go[br]back a page and just show you an

0:16:28.424,0:16:31.864
alternative way of calculating[br]values for the constants A&B in

0:16:31.864,0:16:35.648
the partial fractions, and I[br]want us to return to this

0:16:35.648,0:16:39.776
equation here that we use to[br]find A&B. Let me write that

0:16:39.776,0:16:40.808
equation down again.

0:16:41.490,0:16:47.580
X is equal to[br]a X +3.

0:16:47.580,0:16:50.700
Plus B2 Minus

0:16:50.700,0:16:58.170
X. What I'm going[br]to do is I'm going to

0:16:58.170,0:17:01.420
start by removing the brackets.

0:17:01.650,0:17:05.310
Will have a multiplied by X

0:17:05.310,0:17:10.146
AX. A Times 3[br]which is 3A.

0:17:11.940,0:17:15.258
B times two or two B.

0:17:16.130,0:17:19.346
And be times minus X or[br]minus BX.

0:17:21.050,0:17:24.768
And then what I'm going to do[br]is, I'm going to collect similar

0:17:24.768,0:17:28.200
terms together so you see Ivan[br]Axe here and minus BX there.

0:17:28.770,0:17:33.820
So altogether I have a minus B,[br]lots of X.

0:17:37.280,0:17:41.739
And we've got 3A Plus[br]2B here.

0:17:44.620,0:17:47.890
We now use this equation to[br]equate coefficients on both

0:17:47.890,0:17:52.468
sides. What do we mean by that?[br]Well, what we do is we ask

0:17:52.468,0:17:57.046
ourselves how many X terms do we[br]have on the left and match that

0:17:57.046,0:18:01.624
with the number of X terms that[br]we have on the right. So you

0:18:01.624,0:18:06.529
see, on the left hand side here,[br]if we look at just the ex terms,

0:18:06.529,0:18:13.228
there's 1X. On the right we've[br]got a minus B, lots of X.

0:18:13.260,0:18:18.060
So we've equated the[br]coefficients of X on both sides.

0:18:19.360,0:18:22.948
We can also look at constant[br]terms on both sides. You see the

0:18:22.948,0:18:24.880
three A plus 2B is a constant.

0:18:25.680,0:18:30.356
There are no constant terms on[br]the left, so if we just look at

0:18:30.356,0:18:32.360
constants, there are none on the

0:18:32.360,0:18:37.360
left. And on the right there's[br]3A Plus 2B.

0:18:37.360,0:18:40.690
And you'll see what we have.[br]Here are two simultaneous

0:18:40.690,0:18:44.353
equations for A&B and if we[br]solve these equations we can

0:18:44.353,0:18:48.682
find values for A&B. Let me call[br]that equation one and that one

0:18:48.682,0:18:52.678
equation two. What I'm going to[br]do is I'm going to multiply

0:18:52.678,0:18:57.673
equation one by two so that will[br]end up with two be so that we

0:18:57.673,0:19:01.336
will be able to add these[br]together to eliminate the bees.

0:19:01.336,0:19:06.331
So if I take equation one and I[br]multiply it by two, I'll get 2

0:19:06.331,0:19:13.538
ones or two. 2A minus 2B.[br]Let's call that equation 3.

0:19:14.350,0:19:18.028
If we add equations two and

0:19:18.028,0:19:24.622
three together. We've got 0 + 2,[br]which is 2, three, 8 + 2 A which

0:19:24.622,0:19:30.750
is 5A and two be added to minus[br]2B cancels out, so two is 5. In

0:19:30.750,0:19:35.729
other words, A is 2 over 5,[br]which is the value we had

0:19:35.729,0:19:37.261
earlier on for A.

0:19:37.820,0:19:42.580
We can then take this value for[br]A and substitute it in either of

0:19:42.580,0:19:46.320
these equations and obtain a[br]value for be. So, for example,

0:19:46.320,0:19:48.360
if we substitute in the first

0:19:48.360,0:19:54.790
equation. Will find that one[br]equals a, is 2/5 minus B.

0:19:56.100,0:20:02.049
Rearranging this B is equal to[br]2/5 - 1.

0:20:02.790,0:20:08.782
And 2/5 - 1 is minus 3/5 the[br]same value as we got before.

0:20:09.760,0:20:13.500
So we've seen two ways of[br]finding the values of the

0:20:13.500,0:20:17.580
constants A&B. We can substitute[br]specific values for X or we can

0:20:17.580,0:20:18.940
equate coefficients on both

0:20:18.940,0:20:23.212
sides. Often will need to use a[br]mix of the two methods in order

0:20:23.212,0:20:24.920
to find all the constants in a

0:20:24.920,0:20:31.405
given problem. Let's have[br]a look at

0:20:31.405,0:20:34.456
a definite integral.

0:20:35.170,0:20:39.031
Suppose we want to find the[br]integral from X is one to access

0:20:39.031,0:20:46.560
2. Of three divided by[br]XX plus one with respect to

0:20:46.560,0:20:52.800
X. As before, we examine this[br]integrand and ask ourselves, is

0:20:52.800,0:20:55.175
this a proper or improper

0:20:55.175,0:20:59.045
fraction? Well, the degree of[br]the denominator is too, because

0:20:59.045,0:21:03.010
when we multiply this out, the[br]highest power of X will be 2.

0:21:03.620,0:21:09.236
The degree of the numerator is[br]zero, with really 3X to the zero

0:21:09.236,0:21:12.692
here, so this is an example of a

0:21:12.692,0:21:18.092
proper fraction. On both of[br]these factors are linear

0:21:18.092,0:21:23.960
factors. So as before, I'm going[br]to express the integrand.

0:21:24.580,0:21:27.044
As the sum of its partial[br]fractions. So let's do that

0:21:27.044,0:21:31.040
first of all. 3 divided by XX

0:21:31.040,0:21:35.958
plus one. The appropriate form[br]of partial fractions.

0:21:36.930,0:21:42.750
Are constant. Over the[br]first linear factor.

0:21:42.750,0:21:47.518
Plus another constant over the[br]second linear factor.

0:21:49.160,0:21:52.652
And our job now is to try to[br]find values for A&B.

0:21:53.720,0:22:00.122
We do this by adding these[br]together as we did before,

0:22:00.122,0:22:04.778
common denominator XX plus one[br]in both cases.

0:22:06.810,0:22:10.870
To write a over X as an[br]equivalent fraction with

0:22:10.870,0:22:14.118
this denominator will[br]need to multiply top and

0:22:14.118,0:22:16.148
bottom by X plus one.

0:22:18.090,0:22:23.927
To write B over X plus one with[br]this denominator will need to

0:22:23.927,0:22:26.621
multiply top and bottom by X.

0:22:27.420,0:22:32.840
So now we've given these two[br]fractions a common denominator,

0:22:32.840,0:22:37.718
and we add the fractions[br]together by adding the

0:22:37.718,0:22:43.301
numerators. I'm putting the[br]result over the common

0:22:43.301,0:22:51.224
denominator. So 3 divided by XX[br]plus One is equal to all this.

0:22:53.150,0:22:54.805
The denominators are already the

0:22:54.805,0:23:00.050
same. So we can equate the[br]numerators that gives us the

0:23:00.050,0:23:05.432
equation 3 equals a X plus one[br]plus BX. And this is the

0:23:05.432,0:23:09.986
equation we can use to try to[br]find values for A&B.

0:23:11.260,0:23:13.636
We could equate coefficients, or[br]we can substitute specific

0:23:13.636,0:23:17.332
values for X and what I'm going[br]to do is I'm going to substitute

0:23:17.332,0:23:21.028
the value X is not and the[br]reason why I'm picking X is not

0:23:21.028,0:23:23.404
is because I recognize straight[br]away that's going to.

0:23:23.960,0:23:26.552
Kill off this last term here[br]that'll have gone and will be

0:23:26.552,0:23:28.280
able to just find a value for A.

0:23:29.280,0:23:34.922
So we substitute X is not on the[br]left, will still have 3.

0:23:35.790,0:23:41.601
And on the right we've got not[br]plus one which is one 1A.

0:23:41.630,0:23:43.611
Be times not is not so that

0:23:43.611,0:23:48.248
goes. So In other words, we've[br]got a value for A and a is 3.

0:23:49.640,0:23:54.320
Another sensible value to[br]substitute is X equals minus

0:23:54.320,0:23:56.565
one. Why is that a sensible

0:23:56.565,0:24:00.294
value? Well, that's a sensible[br]value, because if we put X is

0:24:00.294,0:24:03.710
minus one in minus one plus one[br]is zero and will lose this first

0:24:03.710,0:24:05.662
term with the A in and will now

0:24:05.662,0:24:09.814
be. So putting X is minus one[br]will have 3.

0:24:10.360,0:24:14.130
This will become zero and will[br]have be times minus one which is

0:24:14.130,0:24:19.700
minus B. So this tells us that B[br]is actually minus three.

0:24:20.390,0:24:26.366
So now we know the value of a is[br]going to be 3 and B is going to

0:24:26.366,0:24:30.878
be minus three. And the problem[br]of performing this integration

0:24:30.878,0:24:34.244
can be solved by integrating[br]these two terms separately.

0:24:35.630,0:24:41.306
Let's do that now. I'll write[br]these these terms down again.

0:24:41.306,0:24:46.982
We're integrating three over XX[br]plus one with respect to X.

0:24:48.180,0:24:52.310
And we've expressed already this[br]as its partial fractions, and

0:24:52.310,0:24:56.853
found that we're integrating[br]three over X minus three over X,

0:24:56.853,0:25:02.222
plus one with respect to X, and[br]this was a definite integral. It

0:25:02.222,0:25:06.352
had limits on, and the limits[br]will one and two.

0:25:06.890,0:25:10.256
So now we use partial[br]fractions to change this

0:25:10.256,0:25:12.874
algebraic fraction into[br]these two simple integrals.

0:25:13.970,0:25:16.088
Now, these are[br]straightforward to finish

0:25:16.088,0:25:19.618
because the integral of[br]three over X is just three

0:25:19.618,0:25:22.089
natural logarithm of the[br]modulus of X.

0:25:25.650,0:25:27.408
The integral of three over X

0:25:27.408,0:25:32.745
plus one. Is 3 natural logarithm[br]of the modulus of X plus one?

0:25:33.460,0:25:35.800
And there's a minus sign[br]in the middle from that.

0:25:36.860,0:25:40.180
This is a definite

0:25:40.180,0:25:44.110
integral. So we have square[br]brackets and we write the limits

0:25:44.110,0:25:45.660
on the right hand side.

0:25:46.350,0:25:48.890
The problem is nearly[br]finished. All we have to do

0:25:48.890,0:25:50.160
is substitute the limits in.

0:25:51.310,0:25:56.590
Upper limit first when X is[br]2, will have three natural

0:25:56.590,0:25:58.030
log of two.

0:25:59.150,0:26:02.918
When X is 2 in here will have

0:26:02.918,0:26:06.476
minus three. Natural[br]logarithm of 2 + 1, which is

0:26:06.476,0:26:10.727
3. So that's what we get when[br]we put the upper limit in.

0:26:11.750,0:26:16.524
When we put the lower limit in,[br]when X is one will have three

0:26:16.524,0:26:17.888
natural logarithm of 1.

0:26:18.640,0:26:23.398
Minus three natural logarithm of[br]1 + 1, which is 2. So that's

0:26:23.398,0:26:28.156
what we get when we put the[br]lower limiting. And of course we

0:26:28.156,0:26:31.450
want to find the difference of[br]these two quantities.

0:26:32.300,0:26:36.120
Here you'll notice with three

0:26:36.120,0:26:40.634
log 2. And over here[br]there's another three

0:26:40.634,0:26:44.274
log, two with A minus and[br]minus, so we're adding

0:26:44.274,0:26:47.550
another three log 2. So[br]altogether there will be

0:26:47.550,0:26:48.642
6 log 2.

0:26:49.840,0:26:53.220
That's minus three log 3.

0:26:53.220,0:26:56.668
And the logarithm of

0:26:56.668,0:27:00.540
1. Is 0 so that banishes.

0:27:01.430,0:27:04.455
Now we could leave the answer[br]like that, although more often

0:27:04.455,0:27:07.480
than not would probably use[br]the laws of logarithms to try

0:27:07.480,0:27:10.780
to tighten this up a little[br]bit and write it in a

0:27:10.780,0:27:13.255
different way. You should be[br]aware that multiplier outside,

0:27:13.255,0:27:16.830
like this six, can be put[br]inside as a power, so we can

0:27:16.830,0:27:19.580
write this as logarithm of 2[br]to the power 6.

0:27:21.040,0:27:23.960
Subtract again a multiplier[br]outside can move inside as a

0:27:23.960,0:27:28.048
power so we can write this as[br]logarithm of 3 to the power 3.

0:27:29.310,0:27:32.970
And you'll also be aware from[br]your loss of logarithms that if

0:27:32.970,0:27:36.325
we're finding the difference of[br]two logarithms, and we can write

0:27:36.325,0:27:41.205
that as the logarithm of 2 to[br]the power 6 / 3 to the power 3.

0:27:41.450,0:27:44.030
And that's my final answer.

0:27:45.280,0:27:52.309
Let's look at another example in[br]which the denominator contains a

0:27:52.309,0:27:57.421
repeated linear factor. Suppose[br]we're interested in evaluating

0:27:57.421,0:28:05.089
this integral 1 divided by X[br]minus one all squared X Plus

0:28:05.089,0:28:12.118
One, and we want to integrate[br]that with respect to X.

0:28:12.990,0:28:14.862
So again, we have a proper

0:28:14.862,0:28:19.437
fraction. And there is a linear[br]factor here. X plus one, another

0:28:19.437,0:28:23.481
linear factor X minus one. But[br]this is a repeated linear factor

0:28:23.481,0:28:24.829
because it occurs twice.

0:28:25.530,0:28:30.795
The appropriate form of partial[br]fractions will be these.

0:28:31.470,0:28:37.608
We want to[br]constant over the

0:28:37.608,0:28:40.677
linear factor X

0:28:40.677,0:28:47.190
minus one. We want[br]another constant over the linear

0:28:47.190,0:28:52.690
factor repeated X minus one[br]squared. And finally we need

0:28:52.690,0:28:58.190
another constant. See over this[br]linear factor X plus one.

0:28:58.190,0:29:02.040
And our task is before[br]is to try to find values

0:29:02.040,0:29:03.440
for the constants AB&C.

0:29:04.470,0:29:09.042
We do that as before, by[br]expressing each of these over a

0:29:09.042,0:29:12.471
common denominator and the[br]common denominator that we want

0:29:12.471,0:29:15.138
is going to be X minus one.

0:29:15.700,0:29:19.740
Squared X plus[br]one.

0:29:20.780,0:29:24.560
Now to achieve a common[br]denominator of X minus one

0:29:24.560,0:29:29.474
squared X Plus one will need to[br]multiply the top and bottom here

0:29:29.474,0:29:35.144
by X minus 1X plus one. So we[br]have a X minus 1X plus one.

0:29:35.170,0:29:38.455
To achieve the common[br]denominator in this case will

0:29:38.455,0:29:42.105
need to multiply top and bottom[br]by X plus one.

0:29:42.620,0:29:46.554
So we'll have a BX plus one.

0:29:47.120,0:29:51.152
And finally, in this case, to[br]achieve a denominator of X minus

0:29:51.152,0:29:55.856
one squared X Plus one will need[br]to multiply top and bottom by X

0:29:55.856,0:29:56.864
minus 1 squared.

0:29:57.640,0:30:04.376
Now this fraction here[br]is the same as

0:30:04.376,0:30:06.902
this fraction here.

0:30:07.300,0:30:10.479
Their denominators are already[br]the same, so we can equate the

0:30:10.479,0:30:13.947
numerators. So if we just look[br]at the numerators will have one

0:30:13.947,0:30:17.993
on the left is equal to the top[br]line here on the right hand

0:30:17.993,0:30:25.160
side. AX minus[br]1X plus one.

0:30:25.290,0:30:29.209
Plus B. X plus one.

0:30:30.290,0:30:34.110
Plus C. X minus one all

0:30:34.110,0:30:38.720
squared. And now we choose some[br]sensible values for X, so

0:30:38.720,0:30:42.427
there's a lot of these terms[br]will drop away. For example,

0:30:42.427,0:30:46.471
supposing we pick X equals 1,[br]what's the point of picking X

0:30:46.471,0:30:50.178
equals one? Well, if we pick[br]axes one, this first term

0:30:50.178,0:30:51.526
vanishes. We lose a.

0:30:52.220,0:30:55.832
Also, if we pick X equal to 1,[br]the last term vanish is because

0:30:55.832,0:30:59.960
we have a 1 - 1 which is zero[br]and will be just left with the

0:30:59.960,0:31:03.930
term involving be. So by letting[br]XP, one will have one.

0:31:04.610,0:31:07.634
On the left is equal to 0.

0:31:08.200,0:31:12.070
One and one here is 22B.

0:31:12.070,0:31:19.090
And the last term vanishes. In[br]other words, B is equal to 1/2.

0:31:20.380,0:31:24.982
What's another sensible value to[br]pick for X? Well, if we let X

0:31:24.982,0:31:26.044
equal minus one.

0:31:26.780,0:31:30.641
X being minus one will mean that[br]this term vanish is minus one

0:31:30.641,0:31:31.829
plus One is 0.

0:31:32.520,0:31:36.953
This term will vanish minus one,[br]plus one is zero and will be

0:31:36.953,0:31:41.386
able to find see. So I'm going[br]to let X be minus one.

0:31:41.620,0:31:44.707
Will still have the one on the

0:31:44.707,0:31:47.890
left. When X is minus one, this

0:31:47.890,0:31:50.110
goes. This goes.

0:31:50.680,0:31:55.375
And on the right hand side this[br]term here will have minus 1 - 1

0:31:55.375,0:31:59.757
is minus two we square it will[br]get plus four, so will get plus

0:31:59.757,0:32:04.999
4C. In other words, see is going[br]to be 1/4.

0:32:06.070,0:32:11.110
So we've got be. We've got C and[br]now we need to find a value for

0:32:11.110,0:32:15.205
a. Now we can substitute any[br]other value we like in here, so

0:32:15.205,0:32:18.985
I'm actually going to pick X[br]equals 0. It's a nice simple

0:32:18.985,0:32:23.080
value. Effects is zero, will[br]have one on the left if X is

0:32:23.080,0:32:26.230
zero there and there will be[br]left with minus one.

0:32:26.790,0:32:32.404
Minus 1 * 1 is minus 1 - 1[br]times a is minus a.

0:32:33.100,0:32:38.840
Thanks is 0 here. Will just left[br]with B Times one which is be.

0:32:38.850,0:32:44.422
And if X is 0 here will have[br]minus one squared, which is plus

0:32:44.422,0:32:50.392
one plus One Times C Plus C. Now[br]we already know values for B and

0:32:50.392,0:32:55.964
for C, so we substitute these in[br]will have one is equal to minus

0:32:55.964,0:33:00.342
A plus B which is 1/2 plus C[br]which is 1/4.

0:33:01.490,0:33:06.150
So rearranging this will have[br]that a is equal to.

0:33:06.730,0:33:12.260
Well, with a half and a quarter,[br]that's 3/4 and the one from this

0:33:12.260,0:33:17.790
side over the other side is[br]minus one. 3/4 - 1 is going to

0:33:17.790,0:33:18.975
be minus 1/4.

0:33:18.980,0:33:25.972
So there we have our values for[br]a for B and for C, so a being

0:33:25.972,0:33:29.031
minus 1/4 will go back in there.

0:33:29.330,0:33:34.062
Be being a half will go back in[br]here and see being a quarter

0:33:34.062,0:33:38.118
will go back in there and then[br]we'll have three separate pieces

0:33:38.118,0:33:41.498
of integration to do in order to[br]complete the problem.

0:33:42.460,0:33:46.468
Let me write these all down[br]again. I'm going to write the

0:33:46.468,0:33:49.474
integral out and write these[br]three separate ones down.

0:33:49.490,0:33:56.826
So we had that the integral of[br]one over X minus one squared X

0:33:56.826,0:34:03.114
Plus one, all integrated with[br]respect to X, is going to equal.

0:34:03.860,0:34:09.940
Minus 1/4 the integral of one[br]over X minus one.

0:34:10.610,0:34:17.918
They'll be plus 1/2 integral of[br]one over X minus 1 squared.

0:34:18.480,0:34:24.384
And they'll[br]be 1/4

0:34:24.384,0:34:27.336
over X

0:34:27.336,0:34:33.425
plus one.[br]And we want

0:34:33.425,0:34:36.500
to DX in

0:34:36.500,0:34:42.033
every term. So we've[br]used partial fractions to split

0:34:42.033,0:34:45.823
this integrand into three[br]separate terms, and will try and

0:34:45.823,0:34:49.613
finish this off. Now. The first[br]integral straightforward when we

0:34:49.613,0:34:54.161
integrate one over X minus one[br]will end up with just the

0:34:54.161,0:34:57.572
natural logarithm of the[br]denominator, so we'll end up

0:34:57.572,0:35:00.983
with minus 1/4 natural[br]logarithm. The modulus of X

0:35:00.983,0:35:06.208
minus one. To integrate this[br]term, we're integrating one over

0:35:06.208,0:35:10.618
X minus one squared. Let me just[br]do that separately.

0:35:10.770,0:35:16.362
To integrate one over X minus[br]one all squared, we make a

0:35:16.362,0:35:20.090
substitution and let you equals[br]X minus one.

0:35:20.760,0:35:25.282
Do you then will equal du DX,[br]which is just one times DX, so

0:35:25.282,0:35:27.866
do you will be the same as DX?

0:35:28.450,0:35:32.239
The integral will become the[br]integral of one over.

0:35:32.980,0:35:35.676
X minus one squared[br]will be you squared.

0:35:36.820,0:35:38.440
And DX is D.

0:35:39.210,0:35:42.666
Now, this is straightforward to[br]finish because this is

0:35:42.666,0:35:44.970
integrating you to the minus 2.

0:35:45.630,0:35:49.040
Increase the power by one[br]becomes you to the minus 1

0:35:49.040,0:35:50.590
divided by the new power.

0:35:51.090,0:35:52.806
And add a constant of[br]integration.

0:35:54.010,0:35:59.197
So when we do this, integration[br]will end up with minus one over

0:35:59.197,0:36:01.480
you. Which is minus one over.

0:36:02.250,0:36:04.008
X minus one.

0:36:04.660,0:36:10.861
And we can put that back into[br]here now. So the integral half

0:36:10.861,0:36:16.108
integral of one over X minus one[br]squared will be 1/2.

0:36:16.210,0:36:19.876
All that we've got down here,[br]which is minus one over X minus

0:36:19.876,0:36:24.960
one. The constant of integration[br]will add another very end. This

0:36:24.960,0:36:27.318
integral is going to be 1/4.

0:36:27.890,0:36:31.894
The natural logarithm of the[br]modulus of X Plus One and then a

0:36:31.894,0:36:33.742
single constant for all of that.

0:36:35.110,0:36:39.491
Let me just tidy up this little[br]a little bit. The two logarithm

0:36:39.491,0:36:43.198
terms can be combined. We've got[br]a quarter of this logarithm.

0:36:43.198,0:36:46.905
Subtract 1/4 of this logarithm,[br]so together we've got a quarter

0:36:46.905,0:36:50.612
the logarithm. If we use the[br]laws of logarithms, we've gotta

0:36:50.612,0:36:54.656
log subtracted log. So we want[br]the first term X Plus one

0:36:54.656,0:36:56.678
divided by the second term X

0:36:56.678,0:37:03.280
minus one. And then this term[br]can be written as minus 1/2.

0:37:03.280,0:37:06.060
One over X minus one.

0:37:06.060,0:37:09.954
Is a constant of integration[br]at the end, and that's the

0:37:09.954,0:37:10.662
integration complete.

0:37:11.960,0:37:18.440
Now let's look at an example in[br]which we have an improper

0:37:18.440,0:37:21.270
fraction. Suppose we have this

0:37:21.270,0:37:27.474
integral. The[br]degree of

0:37:27.474,0:37:30.730
the numerator

0:37:30.730,0:37:36.332
is 3.[br]The degree of the

0:37:36.332,0:37:37.556
denominator is 2.

0:37:38.610,0:37:41.620
3 being greater than two[br]means that this is an

0:37:41.620,0:37:42.222
improper fraction.

0:37:43.330,0:37:46.386
Improper fractions requires[br]special treatment and the first

0:37:46.386,0:37:49.824
thing we do is division[br]polynomial division. Now, if

0:37:49.824,0:37:53.644
you're not happy with Long[br]Division of polynomials, then I

0:37:53.644,0:37:57.464
would suggest that you look[br]again at the video called

0:37:57.464,0:38:00.868
polynomial division. When[br]examples like this are done very

0:38:00.868,0:38:05.124
thoroughly, but what we want to[br]do is see how many times X minus

0:38:05.124,0:38:06.948
X squared minus four will divide

0:38:06.948,0:38:13.470
into X cubed. The way we do[br]this polynomial division is we

0:38:13.470,0:38:19.630
say how many times does X[br]squared divided into X cubed.

0:38:20.330,0:38:24.878
That's like asking How many[br]times X squared will go into X

0:38:24.878,0:38:29.047
cubed, and clearly when X[br]squared is divided into X cubed,

0:38:29.047,0:38:30.942
the answer is just X.

0:38:31.920,0:38:35.652
So X squared goes into X[br]cubed X times and we write

0:38:35.652,0:38:36.896
the solution down there.

0:38:38.130,0:38:42.186
We take what we have just[br]written down and multiply it by

0:38:42.186,0:38:45.566
everything here. So X times X[br]squared is X cubed.

0:38:46.130,0:38:49.588
X times minus four is minus 4X.

0:38:50.370,0:38:51.878
And then we subtract.

0:38:52.690,0:38:55.469
X cubed minus X cubed vanish is.

0:38:56.040,0:39:00.385
With no access here, and we're[br]subtracting minus 4X, which is

0:39:00.385,0:39:01.570
like adding 4X.

0:39:03.110,0:39:08.141
This means that when we divide X[br]squared minus four into X cubed,

0:39:08.141,0:39:13.559
we get a whole part X and a[br]remainder 4X. In other words, X

0:39:13.559,0:39:18.203
cubed divided by X squared minus[br]four can be written as X.

0:39:19.360,0:39:25.096
Plus 4X divided by X[br]squared minus 4.

0:39:25.270,0:39:30.242
So in order to tackle this[br]integration, we've done the long

0:39:30.242,0:39:34.762
division and we're left with two[br]separate integrals to workout.

0:39:35.640,0:39:37.980
Now this one is going to be[br]straightforward. Clearly you're

0:39:37.980,0:39:41.022
just integrating X, it's going[br]to be X squared over 2. This is

0:39:41.022,0:39:43.596
a bit more problematic. Let's[br]have a look at this again.

0:39:44.420,0:39:48.083
This is now a proper fraction[br]where the degree of the

0:39:48.083,0:39:51.413
numerator is one with the next[br]to the one here.

0:39:52.850,0:39:54.512
The degree of the denominator is

0:39:54.512,0:39:56.810
2. So it's a proper fraction.

0:39:58.130,0:40:00.990
It looks as though we've got a[br]quadratic factor in the

0:40:00.990,0:40:05.690
denominator. But in fact, X[br]squared minus four will

0:40:05.690,0:40:10.420
factorize. It's in fact, the[br]difference of two squares. So we

0:40:10.420,0:40:16.440
can write 4X over X squared[br]minus four as 4X over X minus 2X

0:40:16.440,0:40:20.740
plus two. So the quadratic will[br]actually factorize and you'll

0:40:20.740,0:40:23.320
see. Now we've got two linear

0:40:23.320,0:40:28.483
factors. We can express this in[br]partial fractions. Let's do

0:40:28.483,0:40:35.021
that. We've got 4X over X, minus[br]two X +2, and because each of

0:40:35.021,0:40:40.158
these are linear factors, the[br]appropriate form is going to be

0:40:40.158,0:40:42.493
a over X minus 2.

0:40:43.160,0:40:46.980
Plus B over X +2.

0:40:49.520,0:40:54.990
We add these together using a[br]common dumped common denominator

0:40:54.990,0:41:02.648
X minus two X +2 and will[br]get a X +2 plus BX minus

0:41:02.648,0:41:09.210
2. All over the common[br]denominator, X minus two X +2.

0:41:12.310,0:41:18.988
Now the left hand side and the[br]right hand side of the same. The

0:41:18.988,0:41:23.758
denominators are already the[br]same, so the numerators must be

0:41:23.758,0:41:27.574
the same. 4X must equal a X +2

0:41:27.574,0:41:29.790
plus B. 6 - 2.

0:41:30.850,0:41:35.708
This is the equation that we can[br]use to find the values for A&B.

0:41:37.550,0:41:39.458
Let me write that down again.

0:41:39.990,0:41:45.758
We've got 4X is[br]a X +2.

0:41:46.440,0:41:51.020
Plus BX minus[br]2.

0:41:52.630,0:41:56.844
What's a sensible value to put[br]in for X? Well, if we choose, X

0:41:56.844,0:41:58.951
is 2, will lose the second term.

0:41:59.620,0:42:02.728
So X is 2 is a good[br]value to put in here.

0:42:03.940,0:42:09.100
Faxes 2 on the left hand side[br]will have 4 * 2, which is 8.

0:42:10.070,0:42:15.410
If X is 2, will have 2 +[br]2, which is 44A.

0:42:15.410,0:42:17.940
And this term will vanish.

0:42:19.000,0:42:22.200
So 8 equals 4A.

0:42:22.380,0:42:27.340
A will equal 2 and that's the[br]value for A.

0:42:29.200,0:42:32.684
What's another sensible value to[br]put in for X? Well, if X is

0:42:32.684,0:42:36.436
minus two, will have minus 2 +[br]2, which is zero. Will lose this

0:42:36.436,0:42:44.200
term. So if X is[br]minus two, will have minus 8

0:42:44.200,0:42:46.180
on the left.

0:42:46.190,0:42:47.710
This first term will vanish.

0:42:48.490,0:42:54.490
An ex being minus two here will[br]have minus 2 - 2 is minus four

0:42:54.490,0:42:55.690
so minus 4B.

0:42:55.690,0:43:01.114
So be must also be[br]equal to two.

0:43:01.120,0:43:06.160
So now we've got values for A[br]and for be both equal to two.

0:43:06.160,0:43:08.680
Let's take us back and see what

0:43:08.680,0:43:11.110
that means. It means that when

0:43:11.110,0:43:14.736
we express. This quantity[br]4X over X squared minus

0:43:14.736,0:43:17.868
four in partial fractions[br]like this, the values of

0:43:17.868,0:43:19.956
A&B are both equal to two.

0:43:21.140,0:43:26.740
So the integral were working out[br]is the integral of X +2 over X

0:43:26.740,0:43:29.540
minus 2 + 2 over X +2.

0:43:29.550,0:43:35.694
So there we[br]are. We've now

0:43:35.694,0:43:41.838
got three separate[br]integrals to evaluate,

0:43:41.838,0:43:47.982
and it's straightforward[br]to finish this

0:43:47.982,0:43:54.540
off. The integral of X[br]is X squared divided by two.

0:43:55.650,0:44:02.475
The integral of two over X minus[br]2 equals 2 natural logarithm of

0:44:02.475,0:44:05.625
the modulus of X minus 2.

0:44:06.430,0:44:13.934
And the integral of two over X[br]+2 is 2 natural logarithm of X

0:44:13.934,0:44:16.614
+2 plus a constant of

0:44:16.614,0:44:20.947
integration. If we wanted to do,[br]we could use the laws of

0:44:20.947,0:44:23.617
logarithms to combine these log[br]rhythmic terms here, but I'll

0:44:23.617,0:44:24.952
leave the answer like that.