0:00:00.490,0:00:04.230
In the first of the units on[br]algebraic fractions, we looked

0:00:04.230,0:00:07.970
at what happened when we had a[br]proper fraction with linear

0:00:07.970,0:00:11.030
factors in the denominator of[br]proper fraction with repeated

0:00:11.030,0:00:13.750
linear factors in the[br]denominator, and what happened

0:00:13.750,0:00:18.170
when we had improper fractions.[br]what I want to do in this video

0:00:18.170,0:00:21.910
is look at what happens when we[br]get an irreducible quadratic

0:00:21.910,0:00:24.630
factor when we get an[br]irreducible quadratic factor

0:00:24.630,0:00:28.710
will end up with an integral of[br]something which looks like this

0:00:28.710,0:00:29.730
X plus B.

0:00:30.870,0:00:37.818
Over a X squared plus BX[br]plus, see where the A&B are

0:00:37.818,0:00:41.638
known constants. And this[br]quadratic in the denominator

0:00:41.638,0:00:44.422
cannot be factorized. Now[br]there's various things that

0:00:44.422,0:00:48.946
could happen. It's possible that[br]a could turn out to be 0. Now,

0:00:48.946,0:00:53.122
if it turns out to be 0, what[br]would be left with?

0:00:53.790,0:00:55.490
Is trying to integrate a

0:00:55.490,0:00:57.526
constant. Over this quadratic

0:00:57.526,0:01:04.446
factor. So we'll just end up[br]with a B over AX squared plus BX

0:01:04.446,0:01:09.041
plus C. Now the first example[br]I'd like to show you is what

0:01:09.041,0:01:12.293
we do when we get a situation[br]where we've just got a

0:01:12.293,0:01:15.003
constant on its own, no ex[br]terms over the irreducible

0:01:15.003,0:01:17.713
quadratic factor, so let's[br]have a look at a specific

0:01:17.713,0:01:17.984
example.

0:01:21.240,0:01:24.544
Suppose we want to[br]integrate a constant one.

0:01:25.590,0:01:27.018
Over X squared.

0:01:28.220,0:01:32.310
Plus X plus one. We want[br]to integrate this with

0:01:32.310,0:01:33.537
respect to X.

0:01:35.640,0:01:38.619
This denominator will not[br]factorize if it would factorize,

0:01:38.619,0:01:40.605
would be back to expressing it

0:01:40.605,0:01:44.840
in partial fractions. The way we[br]proceed is to try to complete

0:01:44.840,0:01:46.240
the square in the denominator.

0:01:47.020,0:01:50.440
Let me remind you of how we[br]complete the square for X

0:01:50.440,0:01:51.865
squared plus X plus one.

0:01:52.850,0:01:57.101
It's a complete the square we[br]try to write the first 2 terms.

0:01:58.500,0:02:00.339
As something squared.

0:02:01.260,0:02:04.704
Well, what do we write in[br]this bracket? We want an X

0:02:04.704,0:02:07.861
and clearly when the brackets[br]are all squared out, will get

0:02:07.861,0:02:10.444
an X squared which is that[br]term dealt with.

0:02:11.740,0:02:15.510
To get an ex here, we need[br]actually a term 1/2 here because

0:02:15.510,0:02:19.280
you imagine when you square the[br]brackets out you'll get a half X

0:02:19.280,0:02:22.760
in another half X, which is the[br]whole X which is that.

0:02:24.730,0:02:27.293
We get something we don't want[br]when these brackets are all

0:02:27.293,0:02:28.691
squared out, we'll end it with

0:02:28.691,0:02:32.681
1/2 squared. Which is 1/4 and we[br]don't want a quarter, so I'm

0:02:32.681,0:02:34.247
going to subtract it again here.

0:02:35.540,0:02:38.988
So altogether, all those[br]terms written down there

0:02:38.988,0:02:42.867
are equivalent to the[br]first 2 terms over here.

0:02:44.500,0:02:47.580
And to make these equal, we[br]still need the plus one.

0:02:50.590,0:02:56.200
So tidying this up, we've[br]actually got X plus 1/2 all

0:02:56.200,0:03:02.320
squared, and one subtract 1/4 is[br]3/4. That is the process of

0:03:02.320,0:03:03.850
completing the square.

0:03:12.040,0:03:17.695
OK, how will that help us? Well,[br]it means that what we want to do

0:03:17.695,0:03:21.842
now is considered instead of the[br]integral we started with. We

0:03:21.842,0:03:26.366
want to consider this integral[br]one over X plus 1/2 all squared.

0:03:29.320,0:03:33.533
Plus 3/4 we want to integrate[br]that with respect to X.

0:03:34.430,0:03:37.634
Now, the way I'm going to[br]proceed is going to make a

0:03:37.634,0:03:40.838
substitution in. Here, I'm going[br]to let you be X plus 1/2.

0:03:48.420,0:03:51.876
When we do that, are integral[br]will become the integral of one

0:03:51.876,0:03:55.908
over X plus 1/2 will be just[br]you, so will end up with you

0:03:55.908,0:03:58.628
squared. We've got plus 3/4.

0:04:00.210,0:04:02.106
We need to take care of the DX.

0:04:03.190,0:04:07.450
Now remember that if we want the[br]differential du, that's du DX

0:04:07.450,0:04:12.775
DX. But in this case du DX is[br]just one. This is just one. So

0:04:12.775,0:04:18.100
do you is just DX. So this is[br]nice and simple. The DX we have

0:04:18.100,0:04:19.875
here just becomes a du.

0:04:22.490,0:04:26.384
Now this integral is a standard[br]form. There's a standard result

0:04:26.384,0:04:30.632
which says that if you want to[br]integrate one over a squared

0:04:30.632,0:04:32.756
plus X squared with respect to

0:04:32.756,0:04:38.416
X. That's equal to one over a[br]inverse tangent that's 10 to the

0:04:38.416,0:04:43.498
minus one of X over a plus a[br]constant. Now we will use that

0:04:43.498,0:04:47.491
result to write the answer down[br]to this integral, because this

0:04:47.491,0:04:49.669
is one of these where a.

0:04:50.750,0:04:52.710
Is the square root of 3 over 2?

0:04:53.360,0:04:55.080
That's a squared is 3/4.

0:04:57.320,0:05:03.368
So A is the square root of 3 /[br]2, so we can write down the

0:05:03.368,0:05:08.282
answer to this straight away and[br]this will workout at one over a,

0:05:08.282,0:05:12.818
which is one over root 3 over[br]210 to the minus one.

0:05:14.390,0:05:18.342
Of X over A. In this[br]case it will be U over a

0:05:18.342,0:05:20.166
which is Route 3 over 2.

0:05:21.660,0:05:22.428
Plus a constant.

0:05:24.940,0:05:28.736
Just to tidy this up a little[br]bit where dividing by a fraction

0:05:28.736,0:05:32.240
here. So dividing by Route 3[br]over 2 is like multiplying by

0:05:32.240,0:05:33.408
two over Route 3.

0:05:35.300,0:05:37.030
We've attempted the minus one.

0:05:37.950,0:05:41.430
You we can replace[br]with X plus 1/2.

0:05:45.040,0:05:49.360
And again, dividing by Route 3[br]over 2 is like multiplying by

0:05:49.360,0:05:50.800
two over Route 3.

0:05:52.120,0:05:53.576
And we have a constant[br]at the end.

0:05:56.560,0:05:59.730
And that's the answer. So In[br]other words, to integrate.

0:06:01.000,0:06:03.576
A constant over an[br]irreducible quadratic factor.

0:06:03.576,0:06:07.992
We can complete the square as[br]we did here and then use

0:06:07.992,0:06:10.936
integration by substitution[br]to finish the problem off.

0:06:12.320,0:06:15.807
So that's what happens when[br]we get a constant over the

0:06:15.807,0:06:16.441
quadratic factor.

0:06:20.970,0:06:24.846
What else could happen? It may[br]happen that we get a situation

0:06:24.846,0:06:28.722
like this. We end up with a[br]quadratic function at the bottom

0:06:28.722,0:06:30.660
and it's derivative at the top.

0:06:33.570,0:06:36.558
If that happens, it's very[br]straightforward to finish the

0:06:36.558,0:06:40.210
integration of because we know[br]from a standard result that this

0:06:40.210,0:06:43.530
evaluates to the logarithm of[br]the modulus of the denominator

0:06:43.530,0:06:47.182
plus a constant. So, for[br]example, I'm thinking now of an

0:06:47.182,0:06:48.510
example like this one.

0:06:52.960,0:06:56.768
Again, irreducible quadratic[br]factor in the denominator.

0:06:58.430,0:07:01.436
Attorney X plus be constant[br]times X plus another

0:07:01.436,0:07:04.776
constant on the top and if[br]you inspect this carefully,

0:07:04.776,0:07:08.116
if you look at the bottom[br]here and you differentiate

0:07:08.116,0:07:10.120
it, you'll get 2X plus one.

0:07:11.320,0:07:14.524
So we've got a situation where[br]we've got a function at the

0:07:14.524,0:07:17.728
bottom and it's derivative at[br]the top so we can write this

0:07:17.728,0:07:20.665
down straight away. The answer[br]is going to be the natural

0:07:20.665,0:07:23.068
logarithm of the modulus of[br]what's at the bottom.

0:07:27.740,0:07:31.160
Let's see and that's finished.[br]That's nice and straightforward.

0:07:31.160,0:07:35.340
If you get a situation where[br]you've got something times X

0:07:35.340,0:07:36.480
plus another constant.

0:07:37.070,0:07:39.875
And this top line is not the[br]derivative of the bottom

0:07:39.875,0:07:43.445
line. Then you gotta do a bit[br]more work on it as we'll see

0:07:43.445,0:07:44.465
in the next example.

0:07:48.830,0:07:53.010
Let's have a look at this[br]example. Suppose we want to

0:07:53.010,0:07:58.330
integrate X divided by X squared[br]plus X Plus One, and we want to

0:07:58.330,0:08:00.610
integrate it with respect to X.

0:08:01.940,0:08:05.410
Still, if we differentiate, the[br]bottom line will get 2X.

0:08:06.040,0:08:08.999
Plus One, and that's not[br]what we have at the top.

0:08:08.999,0:08:12.227
However, what we can do is[br]we can introduce it to at

0:08:12.227,0:08:15.455
the top, so we have two X in[br]this following way. By

0:08:15.455,0:08:18.145
little trick we can put a[br]two at the top.

0:08:23.160,0:08:26.046
And in order to make this the[br]same as the integral that we

0:08:26.046,0:08:27.600
started with, I'm going to put a

0:08:27.600,0:08:29.660
factor of 1/2 outside. Half and

0:08:29.660,0:08:33.762
the two canceling. Will will[br]leave the integral that we

0:08:33.762,0:08:34.770
started with that.

0:08:36.010,0:08:39.590
Now. If we differentiate the[br]bottom you see, we get.

0:08:40.140,0:08:44.700
2X. Which is what we've got[br]at the top. But we also get a

0:08:44.700,0:08:46.860
plus one from differentiating[br]the extreme and we haven't

0:08:46.860,0:08:49.500
got a plus one there, so we[br]apply another little trick

0:08:49.500,0:08:50.940
now, and we do the following.

0:08:54.460,0:08:56.716
We'd like a plus one there.

0:09:00.700,0:09:04.120
So that the derivative of[br]the denominator occurs in

0:09:04.120,0:09:04.880
the numerator.

0:09:05.960,0:09:08.914
But this is no longer the same[br]as that because I've added a one

0:09:08.914,0:09:10.391
here. So I've got to take it

0:09:10.391,0:09:13.440
away again. In order that[br]were still with the same

0:09:13.440,0:09:14.540
problem that we started with.

0:09:16.540,0:09:21.258
Now what I can do is I can split[br]this into two integrals. I've

0:09:21.258,0:09:24.628
got a half the integral of these[br]first 2 terms.

0:09:27.300,0:09:29.946
Over X squared plus X plus one.

0:09:31.730,0:09:34.580
DX and I've got a half.

0:09:35.820,0:09:40.236
The integral of the second term,[br]which is minus one over X

0:09:40.236,0:09:45.020
squared plus X plus one DX so[br]that little bit of trickery has

0:09:45.020,0:09:49.068
allowed me to split the thing[br]into two integrals. Now this

0:09:49.068,0:09:52.748
first one we've already seen is[br]straightforward to finish off,

0:09:52.748,0:09:56.428
because the numerator now is the[br]derivative of the denominator,

0:09:56.428,0:09:59.004
so this is just a half the

0:09:59.004,0:10:02.805
natural logarithm. Of the[br]modulus of X squared plus

0:10:02.805,0:10:03.960
X plus one.

0:10:06.620,0:10:10.076
And then we've got minus 1/2.[br]Take the minus sign out minus

0:10:10.076,0:10:13.532
1/2, and this integral integral[br]of one over X squared plus X

0:10:13.532,0:10:17.276
Plus one is the one that we did[br]right at the very beginning.

0:10:17.850,0:10:22.335
And if we just look back, let's[br]see the results of finding that

0:10:22.335,0:10:26.475
integral. Was this one here, two[br]over Route 3 inverse tan of

0:10:26.475,0:10:28.890
twice X plus 1/2 over Route 3?

0:10:29.890,0:10:34.579
So we've got two over Route[br]3 inverse tangent.

0:10:35.750,0:10:37.450
Twice X plus 1/2.

0:10:38.540,0:10:39.728
Over Route 3.

0:10:41.580,0:10:43.010
Plus a constant of integration.

0:10:44.540,0:10:49.340
I can just tidy this up so it's[br]nice and neat to finish it off a

0:10:49.340,0:10:52.340
half the logarithm of X squared[br]plus X plus one.

0:10:54.020,0:10:59.408
The Twos will counsel here, so[br]I'm left with minus one over

0:10:59.408,0:11:01.204
Route 3 inverse tangent.

0:11:01.920,0:11:05.196
And it might be nice just to[br]multiply these brackets out to

0:11:05.196,0:11:08.472
finish it off, so I'll have two[br]X and 2 * 1/2.

0:11:09.110,0:11:09.820
Is one.

0:11:10.890,0:11:12.178
All over Route 3.

0:11:13.800,0:11:15.370
Plus a constant of integration.

0:11:17.670,0:11:19.050
And that's the problem solved.

0:11:26.100,0:11:30.280
Let's have a look at one[br]final example where we can

0:11:30.280,0:11:33.700
draw some of these threads[br]together. Supposing we want

0:11:33.700,0:11:37.500
to integrate 1 divided by XX[br]squared plus one DX.

0:11:39.400,0:11:42.700
What if we got? In this case,[br]it's a proper fraction.

0:11:44.030,0:11:45.482
And we've got a linear factor

0:11:45.482,0:11:47.880
here. And a quadratic[br]factor here.

0:11:48.900,0:11:52.376
You can try, but you'll find[br]that this quadratic factor will

0:11:52.376,0:11:54.272
not factorize, so this is an

0:11:54.272,0:11:57.754
irreducible quadratic factor.[br]So what we're going to do is

0:11:57.754,0:11:59.932
we're going to, first of all,[br]express the integrand.

0:12:02.580,0:12:06.050
As the sum of its partial[br]fractions and the appropriate

0:12:06.050,0:12:09.867
form of partial fractions are[br]going to be a constant over

0:12:09.867,0:12:10.908
the linear factor.

0:12:12.870,0:12:15.586
And then we'll need BX plus C.

0:12:16.850,0:12:20.378
Over the irreducible quadratic[br]factor X squared plus one.

0:12:22.450,0:12:28.255
We now have to find abian. See,[br]we do that in the usual way by

0:12:28.255,0:12:31.738
adding these together, the[br]common denominator will be XX

0:12:31.738,0:12:32.899
squared plus one.

0:12:35.200,0:12:39.208
Will need to multiply top and[br]bottom here by X squared plus

0:12:39.208,0:12:42.882
one to achieve the correct[br]denominator so we'll have an AX

0:12:42.882,0:12:43.884
squared plus one.

0:12:45.530,0:12:51.004
And we need to multiply top and[br]bottom here by X to achieve that

0:12:51.004,0:12:54.914
denominator. So we'll have VX[br]plus C4 multiplied by X.

0:12:57.330,0:12:59.160
This quantity is equal to that

0:12:59.160,0:13:03.143
quantity. The denominators are[br]already the same, so we can

0:13:03.143,0:13:07.303
equate the numerators. If we[br]just look at the numerators

0:13:07.303,0:13:11.252
will have one is equal to a X[br]squared plus one.

0:13:13.650,0:13:15.518
Plus BX Plus C.

0:13:16.820,0:13:18.410
Multiplied by X.

0:13:20.400,0:13:23.777
What's a sensible value to[br]substitute for X so we can

0:13:23.777,0:13:27.154
find abian? See while a[br]sensible value is clearly X is

0:13:27.154,0:13:30.838
zero, whi is that sensible?[br]Well, if X is zero, both of

0:13:30.838,0:13:32.987
these terms at the end will[br]disappear.

0:13:34.520,0:13:38.304
So X being zero will have one is[br]equal to A.

0:13:39.080,0:13:41.078
0 squared is 0 + 1.

0:13:42.080,0:13:45.032
Is still one, so we'll have one[br]a. So a is one.

0:13:46.830,0:13:48.090
That's our value for a.

0:13:49.410,0:13:54.170
What can we do to find B and see[br]what I'm going to do now is I'm

0:13:54.170,0:13:56.690
going to equate some[br]coefficients and let's start by

0:13:56.690,0:13:59.490
looking at the coefficients of X[br]squared on both side.

0:14:01.560,0:14:04.730
On the left hand side there[br]are no X squared's.

0:14:06.350,0:14:09.626
What about on the right[br]hand side? There's clearly

0:14:09.626,0:14:10.718
AX squared here.

0:14:12.990,0:14:15.339
And when we multiply the[br]brackets out here, that

0:14:15.339,0:14:16.383
would be X squared.

0:14:19.960,0:14:24.202
There are no more X squares, so[br]A plus B must be zero. That

0:14:24.202,0:14:28.444
means that B must be the minus[br]negative of a must be the minor

0:14:28.444,0:14:32.080
say, but a is already one, so be[br]must be minus one.

0:14:34.150,0:14:38.518
We still need to find C and[br]will do that by equating

0:14:38.518,0:14:39.610
coefficients of X.

0:14:41.740,0:14:43.484
There are no ex terms on[br]the left.

0:14:45.240,0:14:46.780
There are no ex terms in here.

0:14:48.460,0:14:53.248
There's an X squared term there,[br]and the only ex term is CX, so

0:14:53.248,0:14:54.616
see must be 0.

0:14:56.310,0:14:59.946
So when we express this in its[br]partial fractions, will end up

0:14:59.946,0:15:01.158
with a being one.

0:15:02.720,0:15:06.704
Be being minus one[br]and see being 0.

0:15:09.020,0:15:14.987
So we'll be left with trying to[br]integrate one over X minus X

0:15:14.987,0:15:17.282
over X squared plus one.

0:15:18.990,0:15:19.640
DX

0:15:21.610,0:15:25.294
so we've used partial fractions[br]to split this up into two terms,

0:15:25.294,0:15:28.978
and all we have to do now is[br]completely integration. Let me

0:15:28.978,0:15:30.206
write that down again.

0:15:31.750,0:15:38.386
We want to integrate one over X[br]minus X over X squared plus one

0:15:38.386,0:15:43.600
and all that wants to be[br]integrated with respect to X.

0:15:46.660,0:15:48.430
First term straightforward.

0:15:49.260,0:15:52.388
The integral of one[br]over X is the

0:15:52.388,0:15:54.734
logarithm of the[br]modulus of X.

0:15:56.570,0:15:58.320
To integrate the second term.

0:15:59.040,0:16:03.506
We notice that the numerator is[br]almost the derivative of the

0:16:03.506,0:16:05.942
denominator. If we[br]differentiate, the denominator

0:16:05.942,0:16:07.160
will get 2X.

0:16:08.110,0:16:11.230
There is really only want 1X[br]now. We fiddle that by putting

0:16:11.230,0:16:13.310
it to at the top and a half

0:16:13.310,0:16:17.598
outside like that. So this[br]integral is going to workout

0:16:17.598,0:16:21.538
to be minus 1/2 the[br]logarithm of the modulus of

0:16:21.538,0:16:23.114
X squared plus one.

0:16:24.710,0:16:27.356
And there's a constant of[br]integration at the end.

0:16:29.400,0:16:32.611
And I'll leave the answer like[br]that if you wanted to do. We

0:16:32.611,0:16:34.587
could combine these using the[br]laws of logarithms.

0:16:35.670,0:16:37.861
And that's integration of[br]algebraic fractions. You

0:16:37.861,0:16:41.304
need a lot of practice at[br]that, and there are more

0:16:41.304,0:16:43.182
practice exercises in the[br]accompanying text.