1
00:00:06,880 --> 00:00:07,380
Hi.

2
00:00:07,380 --> 00:00:09,000
Welcome back to recitation.

3
00:00:09,000 --> 00:00:11,710
You've been talking in
class about partial fraction

4
00:00:11,710 --> 00:00:14,670
decomposition as a
tool for integration.

5
00:00:14,670 --> 00:00:17,970
So remember that the point of
partial fraction decomposition

6
00:00:17,970 --> 00:00:21,220
is that whenever you have a
rational function, that is,

7
00:00:21,220 --> 00:00:25,010
one polynomial divided by
another, that in principle,

8
00:00:25,010 --> 00:00:26,830
partial fraction
decomposition lets

9
00:00:26,830 --> 00:00:30,720
you write any such expression as
a sum of things, each of which

10
00:00:30,720 --> 00:00:32,480
is relatively easy to integrate.

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00:00:32,480 --> 00:00:35,800
So the technique here
is purely algebraic.

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00:00:35,800 --> 00:00:37,920
And then you just
apply integral rules

13
00:00:37,920 --> 00:00:39,560
that we've already learned.

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00:00:39,560 --> 00:00:44,892
So I have here four
rational functions for you.

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00:00:44,892 --> 00:00:46,600
And what I'd like you
to do in each case,

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00:00:46,600 --> 00:00:49,670
is try to decompose it
into the general form

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00:00:49,670 --> 00:00:51,800
that Professor
Jerison taught you.

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00:00:51,800 --> 00:00:54,370
So don't, I'm not asking
you to-- if you'd like,

19
00:00:54,370 --> 00:00:56,610
you're certainly welcome
to go ahead and compute

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00:00:56,610 --> 00:00:59,330
the antiderivatives after
you do that, but I'm

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00:00:59,330 --> 00:01:03,150
not going to do it
for you, or I'm not

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00:01:03,150 --> 00:01:04,450
going to ask you to do it.

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00:01:04,450 --> 00:01:05,908
So what I'd like
you to do, though,

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00:01:05,908 --> 00:01:08,560
is for each of these
four expressions,

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00:01:08,560 --> 00:01:12,730
to break it out into the
form of the partial fraction

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00:01:12,730 --> 00:01:13,534
decomposition.

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00:01:13,534 --> 00:01:15,950
So why don't you take a few
minutes to do that, come back,

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00:01:15,950 --> 00:01:17,658
and you can check your
work against mine.

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00:01:26,840 --> 00:01:27,380
All right.

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00:01:27,380 --> 00:01:27,970
Welcome back.

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00:01:27,970 --> 00:01:30,220
Hopefully you had some
fun working on these.

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00:01:30,220 --> 00:01:31,880
They're a little
bit tricky, I think,

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00:01:31,880 --> 00:01:33,920
or I've picked them to
be a little bit tricky.

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00:01:33,920 --> 00:01:35,720
So let's go through
them one by one.

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00:01:35,720 --> 00:01:37,425
I guess I'll start
with the first one.

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00:01:40,710 --> 00:01:49,800
So with the first one, I have
x squared minus 4x plus 4

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00:01:49,800 --> 00:01:53,930
over x squared minus 8x.

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00:01:53,930 --> 00:01:57,360
Now, the first thing to do when
you start the partial fraction

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00:01:57,360 --> 00:01:59,200
method, is that
you have to check

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00:01:59,200 --> 00:02:02,360
that the degree of
the numerator is

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00:02:02,360 --> 00:02:05,190
smaller than the degree
of the denominator.

42
00:02:05,190 --> 00:02:08,520
Now in this case,
that's not true.

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00:02:08,520 --> 00:02:11,720
The degree on top is 2, and
the degree on bottom is 2.

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00:02:11,720 --> 00:02:13,020
So we need to do long division.

45
00:02:13,020 --> 00:02:15,470
Or you know, we need
to do something-- well,

46
00:02:15,470 --> 00:02:18,860
long division is the usual
and always-- usual way,

47
00:02:18,860 --> 00:02:22,400
and always works-- in order to
reduce the degree of the top

48
00:02:22,400 --> 00:02:25,460
here so that it's smaller
than the degree at the bottom.

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00:02:25,460 --> 00:02:27,910
Now, I've done
this ahead of time,

50
00:02:27,910 --> 00:02:31,010
and it's not too hard
to check that this

51
00:02:31,010 --> 00:02:41,290
is equal to 1 plus 4x plus
4 over x squared minus 8x.

52
00:02:41,290 --> 00:02:45,850
It's a relatively easy long
division to do in any case.

53
00:02:45,850 --> 00:02:46,610
So OK.

54
00:02:46,610 --> 00:02:49,260
So what we get after we
do that process is we

55
00:02:49,260 --> 00:02:52,180
get something in front
that's always a polynomial.

56
00:02:52,180 --> 00:02:53,630
And that's good,
because remember,

57
00:02:53,630 --> 00:02:56,620
our goal is to, you know,
manipulate this into a form

58
00:02:56,620 --> 00:02:58,650
where we can integrate
it, and polynomials

59
00:02:58,650 --> 00:02:59,790
are easy to integrate.

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00:02:59,790 --> 00:03:03,230
So then we usually just forget
about this for the time being.

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00:03:03,230 --> 00:03:05,930
And what we want to
do is partial fraction

62
00:03:05,930 --> 00:03:07,850
decompose the second part.

63
00:03:07,850 --> 00:03:10,250
So to do that, the first
thing you need to do,

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00:03:10,250 --> 00:03:13,880
once you've got a smaller
degree on top than downstairs,

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00:03:13,880 --> 00:03:16,740
is that you factor
the denominator.

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00:03:16,740 --> 00:03:20,310
So in this case, I'm
going to keep the 1 plus,

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00:03:20,310 --> 00:03:22,260
just so I can keep
writing equals signs

68
00:03:22,260 --> 00:03:24,060
and be honest about it.

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00:03:24,060 --> 00:03:27,370
But really, our focus
now is just entirely

70
00:03:27,370 --> 00:03:29,420
on this second summand.

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00:03:29,420 --> 00:03:34,000
So this is equal to 1
plus 4x plus 4 is on top.

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00:03:34,000 --> 00:03:37,640
And OK, so we need to factor
the denominator, if possible.

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00:03:37,640 --> 00:03:40,164
And in this case, that's
not only possible.

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00:03:40,164 --> 00:03:41,330
It's pretty straightforward.

75
00:03:41,330 --> 00:03:43,940
We can factor out an
x from both terms,

76
00:03:43,940 --> 00:03:46,660
and we see that
the denominator is

77
00:03:46,660 --> 00:03:51,540
x times x-- whoops-- minus 8.

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00:03:54,290 --> 00:03:55,050
OK.

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00:03:55,050 --> 00:03:58,320
Now, in this case, this
is the simplest situation

80
00:03:58,320 --> 00:03:59,980
for partial fraction
decomposition.

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00:03:59,980 --> 00:04:01,990
We have the denominator
is a product

82
00:04:01,990 --> 00:04:04,110
of distinct linear factors.

83
00:04:04,110 --> 00:04:05,610
So when the denominator
is a product

84
00:04:05,610 --> 00:04:09,430
of distinct linear factors,
what we get this is equal to,

85
00:04:09,430 --> 00:04:12,195
what the partial fraction
theorem tells us we can write,

86
00:04:12,195 --> 00:04:18,100
is that this is equal to
something, some constant,

87
00:04:18,100 --> 00:04:21,240
over the first factor
plus some constant

88
00:04:21,240 --> 00:04:24,490
over the second factor.

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00:04:24,490 --> 00:04:26,980
Now, if you had, you know,
three factors, then you'd

90
00:04:26,980 --> 00:04:30,050
have three of these terms,
one for each factor,

91
00:04:30,050 --> 00:04:32,350
if they were distinct
linear factors.

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00:04:32,350 --> 00:04:33,960
So great.

93
00:04:33,960 --> 00:04:34,620
OK.

94
00:04:34,620 --> 00:04:38,185
Now we can apply
the cover up method

95
00:04:38,185 --> 00:04:39,560
that Professor
Jerison taught us.

96
00:04:39,560 --> 00:04:41,480
So again, the 1 doesn't
really matter here.

97
00:04:41,480 --> 00:04:42,880
You can just ignore it.

98
00:04:42,880 --> 00:04:45,610
So we have this is
equal to this sum,

99
00:04:45,610 --> 00:04:49,770
and we want to find the values
of A and B that make this true.

100
00:04:49,770 --> 00:04:52,720
And then once we have values
of A and B that make this true,

101
00:04:52,720 --> 00:04:55,310
the resulting expression
will be all set to integrate.

102
00:04:55,310 --> 00:04:56,000
Right?

103
00:04:56,000 --> 00:04:57,550
This will just be easy.

104
00:04:57,550 --> 00:05:00,360
It's just a polynomial,
in fact, just a constant.

105
00:05:00,360 --> 00:05:02,460
This is going to
give us a logarithm,

106
00:05:02,460 --> 00:05:04,580
and this is going to
give us a logarithm.

107
00:05:04,580 --> 00:05:07,440
So once we find these
constants A and B,

108
00:05:07,440 --> 00:05:09,407
we're all set to integrate.

109
00:05:09,407 --> 00:05:11,240
So OK, so how does the
cover up method work?

110
00:05:11,240 --> 00:05:15,590
Well, so you cover up
one of the factors.

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00:05:15,590 --> 00:05:17,850
In this case, x.

112
00:05:17,850 --> 00:05:19,390
And you cover up,
on the other side,

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00:05:19,390 --> 00:05:24,680
everything that doesn't have x
in the denominator, and also x.

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00:05:24,680 --> 00:05:26,450
And then you go
back here, and you

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00:05:26,450 --> 00:05:30,420
plug in the appropriate values,
you know, the x minus whatever.

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00:05:30,420 --> 00:05:33,170
So in this case,
that's x equals 0.

117
00:05:33,170 --> 00:05:36,160
And so over here, you
get 4 over minus 8.

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00:05:36,160 --> 00:05:43,890
So that gives us A is equal
to 4 divided by negative 8,

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00:05:43,890 --> 00:05:46,890
which is minus a half.

120
00:05:46,890 --> 00:05:50,300
And now we can do the same
thing with the x minus 8 part.

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00:05:50,300 --> 00:05:53,330
So we cover up x
minus 8, we cover up

122
00:05:53,330 --> 00:05:56,220
everything that doesn't
have an x minus 8 in it.

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00:05:56,220 --> 00:05:58,390
So we've got, on the
right-hand side, we get B,

124
00:05:58,390 --> 00:06:01,080
and on the left hand side,
we have to plug in 8 here.

125
00:06:01,080 --> 00:06:07,650
So we plug 4 times 8 plus 4,
which is 32 plus-- it's 36,

126
00:06:07,650 --> 00:06:10,870
and divided by eight, so
that's 36 divided by 8

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00:06:10,870 --> 00:06:12,480
is 9 divided by 2.

128
00:06:20,820 --> 00:06:23,060
So once you've got these
values of A and B-- OK.

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00:06:23,060 --> 00:06:25,890
So our original
expression, now we

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00:06:25,890 --> 00:06:27,800
substitute in these
values of A and B,

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00:06:27,800 --> 00:06:30,180
then we know that this
is is a true equality,

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00:06:30,180 --> 00:06:33,670
and then the integration
of this expression

133
00:06:33,670 --> 00:06:36,150
is just reduced completely
to the integration

134
00:06:36,150 --> 00:06:38,430
of this easy-to-integrate
expression.

135
00:06:38,430 --> 00:06:38,930
OK?

136
00:06:38,930 --> 00:06:42,020
And so this is the partial
fraction decomposition here,

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00:06:42,020 --> 00:06:46,230
with, you know A equals minus a
half, and B equals nine halves.

138
00:06:46,230 --> 00:06:48,200
So that's part (a).

139
00:06:48,200 --> 00:06:52,060
Let's go on to part (b).

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00:06:52,060 --> 00:06:54,720
I have to remind myself
what part (b) is.

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00:06:54,720 --> 00:06:55,220
OK.

142
00:06:55,220 --> 00:06:58,690
So part (b) is x squared divided
by x plus 2 to the fourth.

143
00:07:07,670 --> 00:07:09,660
So in this case,
in the denominator,

144
00:07:09,660 --> 00:07:12,450
we still have only
linear factors,

145
00:07:12,450 --> 00:07:14,280
but we have repeated factors.

146
00:07:14,280 --> 00:07:17,080
And I mean, this is actually
a particularly simple example,

147
00:07:17,080 --> 00:07:19,960
where we have only one factor,
but it's repeated four times.

148
00:07:19,960 --> 00:07:20,460
Right?

149
00:07:20,460 --> 00:07:21,430
Fourth power.

150
00:07:21,430 --> 00:07:23,790
So when you have that
situation, the partial fraction

151
00:07:23,790 --> 00:07:25,620
decomposition looks a
little bit different.

152
00:07:25,620 --> 00:07:30,130
And so what you get is that you
have, for every repeated power,

153
00:07:30,130 --> 00:07:33,070
so for each-- this one
appears four times,

154
00:07:33,070 --> 00:07:36,130
so you get four summands
on the right-hand side,

155
00:07:36,130 --> 00:07:39,180
with increasing powers
of this linear factor.

156
00:07:39,180 --> 00:07:41,420
So with increasing
powers of x minus 1.

157
00:07:41,420 --> 00:07:46,920
So this is going to be
A-- or sorry, x plus 1.

158
00:07:46,920 --> 00:07:53,960
A over x plus 1
plus B over x plus 1

159
00:07:53,960 --> 00:08:04,890
squared plus C over x plus
1 cubed plus D over x plus 1

160
00:08:04,890 --> 00:08:05,560
to the fourth.

161
00:08:05,560 --> 00:08:08,330
Now remember, even
though the degree here

162
00:08:08,330 --> 00:08:13,690
goes up in these later summands,
what stays on top is the same.

163
00:08:13,690 --> 00:08:15,460
It just stays a constant.

164
00:08:15,460 --> 00:08:18,500
If this were a quadratic
factor, it would stay linear.

165
00:08:18,500 --> 00:08:22,430
The top doesn't increase
in degree when do this.

166
00:08:22,430 --> 00:08:25,920
And a good simple way
to check that you've

167
00:08:25,920 --> 00:08:29,646
got the right-- you know, before
you solve for the constants,

168
00:08:29,646 --> 00:08:32,020
make sure you've got the
correct abstract decomposition--

169
00:08:32,020 --> 00:08:36,890
is to count the number
of these constants

170
00:08:36,890 --> 00:08:38,170
that you're looking for.

171
00:08:38,170 --> 00:08:41,380
It should always
match the degree

172
00:08:41,380 --> 00:08:44,280
of the denominator over here.

173
00:08:44,280 --> 00:08:47,492
So in this case, the degree
of the denominator is 4,

174
00:08:47,492 --> 00:08:48,575
and there are 4 constants.

175
00:08:48,575 --> 00:08:52,760
So we've-- so that's a good
way to check that you set

176
00:08:52,760 --> 00:08:54,410
the problem up right.

177
00:08:54,410 --> 00:08:54,940
OK.

178
00:08:54,940 --> 00:08:57,020
So now, what do we
do in this case?

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00:08:57,020 --> 00:08:59,830
Well, the cover up
method works, but it only

180
00:08:59,830 --> 00:09:02,590
works to find the
highest degree term.

181
00:09:02,590 --> 00:09:03,090
Right?

182
00:09:03,090 --> 00:09:06,830
So we cover up x
plus 1 to the fourth,

183
00:09:06,830 --> 00:09:11,940
and we cover up everything with
a smaller power of x plus 1,

184
00:09:11,940 --> 00:09:14,880
and then we plug in negative 1.

185
00:09:14,880 --> 00:09:15,380
Right?

186
00:09:15,380 --> 00:09:17,730
Because we need
x plus 1 to be 0.

187
00:09:17,730 --> 00:09:21,210
So OK, so over here we get
negative 1 squared is 1,

188
00:09:21,210 --> 00:09:23,750
so we get, right away,
that D is equal to 1.

189
00:09:27,510 --> 00:09:28,010
OK.

190
00:09:28,010 --> 00:09:30,460
But that doesn't
give us A, B, or C.

191
00:09:30,460 --> 00:09:34,860
We can't get A, B, or C
by the cover up method.

192
00:09:34,860 --> 00:09:36,510
Now, there are a
couple different ways

193
00:09:36,510 --> 00:09:38,730
you can proceed at this point.

194
00:09:38,730 --> 00:09:43,910
One thing you can do, which
is something that often works,

195
00:09:43,910 --> 00:09:47,230
is you could plug in values.

196
00:09:47,230 --> 00:09:48,660
Well, so this will always work.

197
00:09:48,660 --> 00:09:50,000
I shouldn't say often works.

198
00:09:50,000 --> 00:09:53,200
You can start plugging
in other values for x.

199
00:09:53,200 --> 00:09:56,060
And as you plug in other
values for x, what you'll see

200
00:09:56,060 --> 00:09:58,050
is that for every value
you plug in, you'll

201
00:09:58,050 --> 00:10:01,890
get a linear equation
relating your variables-- A,

202
00:10:01,890 --> 00:10:04,640
B, C, and whatever,
in this case, that's

203
00:10:04,640 --> 00:10:07,250
all we've got left, A,
B, and C-- to each other.

204
00:10:07,250 --> 00:10:10,630
And so if you plug in three
different values of x, say,

205
00:10:10,630 --> 00:10:12,470
you'll get three
different linear equations

206
00:10:12,470 --> 00:10:15,230
with three different variables,
and then you can solve them.

207
00:10:15,230 --> 00:10:17,430
That's one thing you can do.

208
00:10:17,430 --> 00:10:21,500
Another thing you could do is
you can multiply through by x

209
00:10:21,500 --> 00:10:23,335
plus 1 to the fourth.

210
00:10:23,335 --> 00:10:25,210
So if you do that, you'll
have-- on the left,

211
00:10:25,210 --> 00:10:27,980
you'll just have x
squared, and on the right

212
00:10:27,980 --> 00:10:30,280
you'll have-- well,
you'll have A times

213
00:10:30,280 --> 00:10:34,850
x plus 1 cubed plus B times x
plus 1 squared, plus C times

214
00:10:34,850 --> 00:10:39,070
x plus 1, plus D. And we already
know that x is equal to 1.

215
00:10:39,070 --> 00:10:39,570
OK?

216
00:10:39,570 --> 00:10:42,320
And so then, for those
two things to be equal,

217
00:10:42,320 --> 00:10:44,000
they're equal as
polynomials, all

218
00:10:44,000 --> 00:10:46,270
their coefficients
have to be equal.

219
00:10:46,270 --> 00:10:49,120
So you can just look at the
coefficients in that resulting

220
00:10:49,120 --> 00:10:56,500
expression, and ask, you know,
which coefficients-- sorry.

221
00:10:56,500 --> 00:10:58,700
You can set coefficients
on the two sides equal.

222
00:10:58,700 --> 00:11:00,174
The two polynomials
are equal, all

223
00:11:00,174 --> 00:11:02,090
of their corresponding
coefficients are equal.

224
00:11:02,090 --> 00:11:03,730
So you could look,
you know, at this side

225
00:11:03,730 --> 00:11:04,688
and you'll say, oh, OK.

226
00:11:04,688 --> 00:11:07,004
So the coefficient
of x cubed has

227
00:11:07,004 --> 00:11:09,420
to be the same as whatever the
coefficient of x cubed over

228
00:11:09,420 --> 00:11:11,876
here is, and the
coefficient x squared

229
00:11:11,876 --> 00:11:14,000
has to be the same as the
coefficient of x squared,

230
00:11:14,000 --> 00:11:14,920
and so on.

231
00:11:14,920 --> 00:11:17,010
So that's another
way to proceed.

232
00:11:24,690 --> 00:11:25,550
Yeah, all right.

233
00:11:25,550 --> 00:11:29,260
Those are really your
two best options.

234
00:11:29,260 --> 00:11:33,110
I like to multiply
through, personally.

235
00:11:33,110 --> 00:11:33,890
So OK.

236
00:11:33,890 --> 00:11:36,880
So if I were to do that, in
this case, on the left-hand side

237
00:11:36,880 --> 00:11:44,560
I'd get x squared equals
A times x plus 1 cubed

238
00:11:44,560 --> 00:11:54,470
plus B times x plus 1 squared
plus C times x plus 1 plus D.

239
00:11:54,470 --> 00:11:56,360
Except we already know
that D is equal to 1,

240
00:11:56,360 --> 00:11:58,670
so I'm just going
to write plus 1.

241
00:12:01,700 --> 00:12:02,230
So OK.

242
00:12:02,230 --> 00:12:04,280
So actually, let
me say that there

243
00:12:04,280 --> 00:12:05,610
are other things you could do.

244
00:12:05,610 --> 00:12:07,460
Which is, you could
rearrange things

245
00:12:07,460 --> 00:12:11,070
and simplify algebraically
before plugging in values,

246
00:12:11,070 --> 00:12:14,430
or before comparing
coefficients.

247
00:12:14,430 --> 00:12:18,210
So let me give you
one example of each

248
00:12:18,210 --> 00:12:19,800
of those three possibilities.

249
00:12:19,800 --> 00:12:22,390
So for example, one
thing you can do,

250
00:12:22,390 --> 00:12:25,430
is you can look at the
highest degree coefficient.

251
00:12:25,430 --> 00:12:27,770
So as Professor Jerison said,
the easiest coefficients

252
00:12:27,770 --> 00:12:30,889
are usually the high-order
terms and the low-order terms.

253
00:12:30,889 --> 00:12:32,430
So in this case,
the high-order terms

254
00:12:32,430 --> 00:12:34,534
would be-- this
is a third degree

255
00:12:34,534 --> 00:12:35,950
polynomial on the
right-hand side,

256
00:12:35,950 --> 00:12:38,180
and it's a second degree
polynomial on the left.

257
00:12:38,180 --> 00:12:40,590
So the highest-order
term here, x cubed,

258
00:12:40,590 --> 00:12:43,700
just appears in this one
place as coefficient A. Right?

259
00:12:43,700 --> 00:12:46,940
This is A x cubed
plus something times x

260
00:12:46,940 --> 00:12:48,630
squared plus blah blah blah.

261
00:12:48,630 --> 00:12:51,640
And over here, we
have no x cubeds.

262
00:12:51,640 --> 00:12:53,889
So we have x cubeds here,
but no x cubeds here.

263
00:12:53,889 --> 00:12:56,180
That means the coefficient
of x cubed here has to be 0,

264
00:12:56,180 --> 00:12:57,120
so A has to be 0.

265
00:13:01,160 --> 00:13:02,050
OK.

266
00:13:02,050 --> 00:13:05,089
So A has to be equal to 0,
and that simplifies everything

267
00:13:05,089 --> 00:13:05,630
a little bit.

268
00:13:05,630 --> 00:13:11,780
So now we get x squared equals
B times x plus 1 squared

269
00:13:11,780 --> 00:13:18,310
plus C times x plus 1 plus 1.

270
00:13:18,310 --> 00:13:23,630
Now let me show you what I mean
about algebraic manipulation.

271
00:13:23,630 --> 00:13:26,570
This 1, if you wanted, you could
always just subtract it over

272
00:13:26,570 --> 00:13:27,660
to the other side.

273
00:13:27,660 --> 00:13:28,190
Right?

274
00:13:28,190 --> 00:13:29,689
And so then you'll
have, on the left

275
00:13:29,689 --> 00:13:31,390
you'll have x squared minus 1.

276
00:13:31,390 --> 00:13:33,440
And x squared minus
1, you can write

277
00:13:33,440 --> 00:13:42,480
as x minus 1 times x plus 1
equals B times x plus 1 squared

278
00:13:42,480 --> 00:13:46,545
plus C times x plus 1.

279
00:13:46,545 --> 00:13:48,795
And then you can divide out
by an x plus 1 everywhere.

280
00:13:48,795 --> 00:13:50,530
It appears in all terms.

281
00:13:50,530 --> 00:13:59,570
So you get x minus 1 equals
B times x plus 1 plus C.

282
00:13:59,570 --> 00:14:02,460
And now what this does for you,
is you sort of just reduced

283
00:14:02,460 --> 00:14:03,400
the degree everywhere.

284
00:14:03,400 --> 00:14:06,210
And actually, you could
substitute x equals minus 1

285
00:14:06,210 --> 00:14:09,960
again, if you wanted
to, for example.

286
00:14:09,960 --> 00:14:13,650
And here, so for example, if
you substitute x equals minus 1,

287
00:14:13,650 --> 00:14:18,330
that's the same idea as what
you do in the cover up method.

288
00:14:18,330 --> 00:14:20,780
This B term will
just die completely,

289
00:14:20,780 --> 00:14:22,860
and you'll be left with
negative 2 on the left.

290
00:14:22,860 --> 00:14:26,780
So you get C-- I'm going to
have to move over here, sorry.

291
00:14:26,780 --> 00:14:28,540
C is equal to negative 2, then.

292
00:14:28,540 --> 00:14:31,380
And also, you can
do the one thing

293
00:14:31,380 --> 00:14:33,280
that I haven't done so
far, is this plugging

294
00:14:33,280 --> 00:14:35,130
in nice choices of values.

295
00:14:35,130 --> 00:14:37,980
So another nice choice of value
for x that we haven't used

296
00:14:37,980 --> 00:14:39,370
is x equals 0.

297
00:14:39,370 --> 00:14:42,230
So if you plug in x equals
0, you'll get minus 1

298
00:14:42,230 --> 00:14:44,815
equals B plus c.

299
00:14:44,815 --> 00:14:50,310
So minus 1 equals B plus C.
And since we just found C,

300
00:14:50,310 --> 00:14:54,760
that means that B is equal to 1.

301
00:14:54,760 --> 00:14:55,260
All right.

302
00:14:55,260 --> 00:14:59,200
So-- oh boy, I'm using a lot
of space, aren't I. All right.

303
00:14:59,200 --> 00:15:02,400
So in this case, we've got
our coefficients, A equals 0,

304
00:15:02,400 --> 00:15:05,580
D equals 1, B equals
1, C equals minus 2.

305
00:15:05,580 --> 00:15:09,400
And that gives us the partial
fraction decomposition.

306
00:15:09,400 --> 00:15:15,770
Let's go back over here then,
and look at question (c).

307
00:15:15,770 --> 00:15:19,080
So for (c), the question is,
what is the partial fraction

308
00:15:19,080 --> 00:15:23,390
decomposition of 2x plus
2 divided by the quantity

309
00:15:23,390 --> 00:15:27,492
4x squared plus 1 squared?

310
00:15:27,492 --> 00:15:28,450
This one's really easy.

311
00:15:28,450 --> 00:15:29,880
This one is done.

312
00:15:29,880 --> 00:15:32,190
This is already partial
fraction decomposed.

313
00:15:32,190 --> 00:15:32,690
Right?

314
00:15:32,690 --> 00:15:36,880
When you have-- so here we
have an irreducible quadratic

315
00:15:36,880 --> 00:15:37,950
in the denominator.

316
00:15:37,950 --> 00:15:42,400
You can't factor this any
further than it's gone.

317
00:15:42,400 --> 00:15:44,780
It also occurs to
a higher power.

318
00:15:44,780 --> 00:15:47,680
So when you partial fraction
decompose something like this,

319
00:15:47,680 --> 00:15:52,010
you want something linear
over 4x squared plus 1,

320
00:15:52,010 --> 00:15:56,760
plus something linear over
4x squared plus 1 squared.

321
00:15:56,760 --> 00:15:58,690
But we already have that, right?

322
00:15:58,690 --> 00:16:01,860
The first linear part
is 0, and the second

323
00:16:01,860 --> 00:16:05,490
is something linear over
4x squared plus 1 squared.

324
00:16:05,490 --> 00:16:08,040
So to integrate this, it's
already in a pretty good form.

325
00:16:08,040 --> 00:16:09,540
Now, you're actually
going to write,

326
00:16:09,540 --> 00:16:11,039
if you wanted to
integrate this, you

327
00:16:11,039 --> 00:16:14,510
would split it into two
pieces, one with the 2x

328
00:16:14,510 --> 00:16:15,900
and then one with the 2.

329
00:16:15,900 --> 00:16:18,040
And the first one,
you would just

330
00:16:18,040 --> 00:16:20,760
be a usual u substitution,
and the second one,

331
00:16:20,760 --> 00:16:24,190
we would want some sort of
trigonometric substitution.

332
00:16:24,190 --> 00:16:28,170
But this one is already
ready for methods we already

333
00:16:28,170 --> 00:16:29,430
should be comfortable with.

334
00:16:29,430 --> 00:16:29,930
OK.

335
00:16:29,930 --> 00:16:31,090
So C, that's easy.

336
00:16:31,090 --> 00:16:32,255
It's done already.

337
00:16:32,255 --> 00:16:35,650
I'll put a check mark there,
because that makes me happy.

338
00:16:35,650 --> 00:16:36,160
OK.

339
00:16:36,160 --> 00:16:39,460
And for-- all right, and
so for this last one,

340
00:16:39,460 --> 00:16:42,470
I'm also not going to
write this one out.

341
00:16:42,470 --> 00:16:45,059
But the thing to notice
here is the way I wrote it--

342
00:16:45,059 --> 00:16:46,350
and this was really mean of me.

343
00:16:46,350 --> 00:16:46,890
Right?

344
00:16:46,890 --> 00:16:49,630
I wrote it as x squared
minus 1 quantity squared.

345
00:16:49,630 --> 00:16:52,240
So a natural instinct
is to say, aha!

346
00:16:52,240 --> 00:16:54,840
It's a quadratic repeated
factor in the denominator.

347
00:16:54,840 --> 00:16:55,340
Right?

348
00:16:55,340 --> 00:16:58,490
But that's just because I was
mean and I wrote it this way.

349
00:16:58,490 --> 00:16:59,870
That's not actually
what this is.

350
00:16:59,870 --> 00:17:01,810
This is not irreducible.

351
00:17:01,810 --> 00:17:02,760
This factors.

352
00:17:02,760 --> 00:17:07,723
You can rewrite-- let
me come back over here.

353
00:17:07,723 --> 00:17:11,730
This is for question (d).

354
00:17:11,730 --> 00:17:18,520
So you can rewrite x squared
minus 1 squared as x minus 1

355
00:17:18,520 --> 00:17:22,100
squared times x plus 1 squared.

356
00:17:22,100 --> 00:17:27,190
You can factor this
x squared minus 1.

357
00:17:27,190 --> 00:17:29,430
So when you factor this
x minus 1, what you see

358
00:17:29,430 --> 00:17:31,440
is-- this isn't a
problem that has

359
00:17:31,440 --> 00:17:36,230
one irreducible quadratic factor
appearing to the second power.

360
00:17:36,230 --> 00:17:38,850
What it has is two
linear factors,

361
00:17:38,850 --> 00:17:40,840
each appearing to
the second power.

362
00:17:40,840 --> 00:17:43,700
So the partial fraction
decomposition in this problem

363
00:17:43,700 --> 00:17:49,800
will be something
like A over x minus 1

364
00:17:49,800 --> 00:17:58,320
plus B over x minus 1
squared plus C over x plus 1,

365
00:17:58,320 --> 00:18:03,180
plus D over x plus 1 squared.

366
00:18:03,180 --> 00:18:06,110
That's what you'll get when
you apply partial fraction

367
00:18:06,110 --> 00:18:07,450
decomposition to this problem.

368
00:18:07,450 --> 00:18:11,030
And then you have to solve
for the coefficients A, B, C,

369
00:18:11,030 --> 00:18:11,530
and D.

370
00:18:11,530 --> 00:18:15,490
So I'm not going to write that
out myself, but I cleverly

371
00:18:15,490 --> 00:18:18,644
did it before I
came on camera, so I

372
00:18:18,644 --> 00:18:21,310
can tell you what the answer is,
if you want to check your work.

373
00:18:21,310 --> 00:18:30,740
So here we have A is equal
to 0, B is equal to 1,

374
00:18:30,740 --> 00:18:35,690
C is equal to 1, and
D is equal to minus 3.

375
00:18:35,690 --> 00:18:38,300
So that's for the-- I
didn't write it over here.

376
00:18:38,300 --> 00:18:40,720
That's for this
particular numerator

377
00:18:40,720 --> 00:18:43,950
that we started
with back over here.

378
00:18:43,950 --> 00:18:46,690
So for this particular
fraction, if you carry out

379
00:18:46,690 --> 00:18:50,970
the partial fraction
decomposition, what you get

380
00:18:50,970 --> 00:18:53,810
is right here.

381
00:18:53,810 --> 00:18:56,170
So OK, so those are, that
was a few more examples

382
00:18:56,170 --> 00:18:58,210
of the partial
fraction decomposition.

383
00:18:58,210 --> 00:19:02,109
I hope you enjoyed them,
and I'm going to end there.