WEBVTT
1
00:00:00.000 --> 00:00:05.040
to what procedure? Partial fractions. That's right. Partial fraction decomposition.
2
00:00:05.040 --> 00:00:10.880
Now, here comes the question, is that quadratic factor ball? Well, you can some of you
3
00:00:10.880 --> 00:00:15.600
can use the guess and check method. But how do you know if something's factor ball?
4
00:00:15.600 --> 00:00:21.360
Well, realistic, if we go back to beginning algebra, if we can find factors of A times C that
5
00:00:21.360 --> 00:00:26.400
some to be, it is factor ball. And so, Charlie, what's A times C with this? If we're looking at AX
6
00:00:26.400 --> 00:00:31.680
squared plus BX plus C, what's A times C? Negative 9. Negative 9, that's right. And what are the
7
00:00:31.680 --> 00:00:36.240
factors of negative 9 that some to a positive 8? Nine and negative 1. A nine and a negative 1,
8
00:00:36.240 --> 00:00:41.360
it is factor ball. So, you should know it's factor ball. Okay, Charlie. And so, you can use the
9
00:00:41.360 --> 00:00:45.280
guess or check method, whatever you want. But how is the factor, Charlie? Three U minus 1.
10
00:00:45.280 --> 00:00:52.800
That's right. U plus 3. Plus 3. Very good, D U. So now, we finally got it factor. So, what are we
11
00:00:52.800 --> 00:00:57.840
going to do now, Charlie? Partial fraction D. Partial fraction D composition. That's right. Okay, so,
12
00:00:58.480 --> 00:01:06.000
here we go. Well, Charlie, we got a long way to go. All right, so take a quick break here.
13
00:01:08.000 --> 00:01:12.000
Okay, we're ready. All right, here we go, Charlie. Partial fraction D composition. We're going to
14
00:01:12.000 --> 00:01:17.680
take our rational function there. And we're going to decompose it into fractions with the
15
00:01:17.680 --> 00:01:24.000
denominators of 3 U minus 1 and U plus 3. Now, remember, when you do partial fraction D composition,
16
00:01:24.000 --> 00:01:28.960
if your denominator is a first degree, then your numerator must be 1 less degree, which in this
17
00:01:28.960 --> 00:01:33.760
case is a zero. So, we'll put 8 represent that constant. Remember, a zero degree polynomial is
18
00:01:33.760 --> 00:01:39.760
a constant. Similarly, with the U plus 3, the constant above should be B. We'll let that be B.
19
00:01:39.760 --> 00:01:45.360
Now, our goal is to solve for A and B, right, Charlie? Okay, so, how are we going to do that?
20
00:01:45.360 --> 00:01:50.480
Multiply both sides by the LCD. We're going to multiply both sides by the lowest common denominator
21
00:01:50.480 --> 00:01:55.520
to clear those fractions out, or sometimes we like to say, come through those fractions.
22
00:01:57.520 --> 00:02:03.520
Anyway, okay, so what is our lowest common denominator, Charlie? 3 U minus 1 times U plus 3.
23
00:02:03.520 --> 00:02:08.560
That's right. 3 U minus 1 times U plus 3. Very nice. So, when we multiply both sides,
24
00:02:08.560 --> 00:02:15.200
what do we get left hand side, Charlie? 2. 2, that's right. And then we have A times what? U plus 3.
25
00:02:15.760 --> 00:02:22.000
U plus 3. Very nice, Charlie. And finally, we have B times 3 U minus 1. 3 U minus 1. Very nice.
26
00:02:22.000 --> 00:02:29.280
Okay, here we go, Charlie. Now, we have to solve for A and B. Now, remember, there's two techniques
27
00:02:29.280 --> 00:02:36.000
we can use. We can use the elimination method by choosing values of U that will eliminate the A or B,
28
00:02:36.000 --> 00:02:40.960
or you can equate terms. Now, this one isn't that tough. We're going to go ahead and eliminate terms
29
00:02:40.960 --> 00:02:45.760
by choosing values of U. So, here we go, Charlie. So, what we're going to first do is we're going to
30
00:02:45.760 --> 00:02:51.440
eliminate that A by choosing U equal what, Charlie? Negative 3. Negative 3, that's right. And so,
31
00:02:51.440 --> 00:02:57.680
when we plug that in, we get 2 equals B times what, Charlie? When you plug in negative 3 there,
32
00:02:57.680 --> 00:03:02.960
remember, it's 3 times U. So, Charlie, it's 1. What do you get? Negative 10. Negative 10, very nice,
33
00:03:02.960 --> 00:03:09.520
Charlie. So, now, solving for B here, okay, what do we get? Negative 1-5. Negative 1-5. So, we have
34
00:03:09.520 --> 00:03:16.880
our B value now. Okay, now, now, done, Charlie. We've got to go for the A value. In order to get the
35
00:03:16.880 --> 00:03:23.120
A value, we have to eliminate that B. Now, look at the B. B is 3 U, subtract 1. What do we have to
36
00:03:23.120 --> 00:03:32.000
choose for U to eliminate that B? 1-3. 1-3, that's right. Because 3 times 1-3 will be 1 and 1-3-1 is 0.
37
00:03:32.000 --> 00:03:37.920
So, if we do that, Charlie, left hand side, we get 2 and then we have A times U plus 3.
38
00:03:39.680 --> 00:03:49.120
Now, Charlie, if U is 1-3, how much is 1-3 plus 3? Remember, 3 is 9-3-3. 10-3-3 is very nice, Charlie.
39
00:03:49.120 --> 00:03:56.160
So, A is equal to, solve for A. What do you get, Charlie? Six tenths, three fifths, three fifths.
40
00:03:56.160 --> 00:04:02.240
That's right, you get six tenths, which reduces to three fifths. All right, Charlie, we finally got
41
00:04:02.240 --> 00:04:10.080
our A and B and now, we can go ahead and take our integral and decompose our rational function
42
00:04:10.080 --> 00:04:16.240
into two separate fractions, right? And hopefully, we can integrate. All right, here we go. So,
43
00:04:16.240 --> 00:04:28.480
remember, we had A over 3U minus 1, right? And so, for the first one, what was A equal to Charlie?
44
00:04:28.480 --> 00:04:33.520
Three fifths. Three fifths, that's right. A was three fifths. And so now, we're going to take the
45
00:04:33.520 --> 00:04:39.040
constant three fifths out of the integral and write this as 1 over 3U minus 1. Because remember,
46
00:04:39.040 --> 00:04:43.760
it was A over 3U minus 1, which is three fifths over 3U minus 1. So, you can take the three fifths
47
00:04:43.760 --> 00:04:49.840
out. Okay, now, let's go to our B term. What was B, Charlie? Negative one fifth, negative one fifth,
48
00:04:49.840 --> 00:04:55.440
and that would take the one fifth out as a constant, and we have integral of 1 over U plus 3U.
49
00:04:56.320 --> 00:05:02.960
All right, Charlie. So, now, we're going to use a technique here to integrate this three fifths
50
00:05:02.960 --> 00:05:09.200
integral of 1 over 3U minus 1. You could do it by doing another substitution by letting V
51
00:05:09.200 --> 00:05:15.280
equal 3U minus 1. But we can avoid that here. Watch. Okay, what I'm going to do, Charlie, is take
52
00:05:15.760 --> 00:05:20.640
from the three fifths. I'm going to take that three and put it on top of the three U minus 1,
53
00:05:20.640 --> 00:05:26.320
just like that. Because this puts this right into the form of a natural log integral.
54
00:05:26.320 --> 00:05:33.280
Because this integral of 3 over 3U minus 1, DU, is the natural log of the absolute value of 3U
55
00:05:33.280 --> 00:05:39.600
minus 1. This is because, notice here, in the denominator, we have 3U minus 1, Charlie. What's a
56
00:05:39.600 --> 00:05:47.520
derivative of 3U minus 1? 3DU. 3DU. So, it's like integrating 1 over U. Okay, and so, that is a
57
00:05:47.520 --> 00:05:52.080
natural log integral. Now, the second one, we'll go ahead and write that down. That is also a natural
58
00:05:52.080 --> 00:05:58.640
log, because the denominator is U plus 3, and the derivative of U plus 3 is 1DU. So, they're right
59
00:05:58.640 --> 00:06:03.600
there, natural log integral forms. Okay, so here we go, Charlie, we're almost home.
60
00:06:05.440 --> 00:06:12.240
All right, here we go. We have 1 fifth. Now, what's the integral of 3 over 3U minus 1, DU?
61
00:06:12.880 --> 00:06:18.240
Natural log of the absolute value of 3U minus 1. Very nice, Charlie. Now,
62
00:06:19.760 --> 00:06:27.200
what's the integral of 1 over U plus 3? Natural log of U plus 3, and that's a 1 fifth. And
63
00:06:27.200 --> 00:06:35.120
don't forget, we have to put a white here, Charlie. C plus C, that's right. Now, we're not done yet.
64
00:06:36.240 --> 00:06:44.640
That's right. Yeah, this is a long problem. Actually, this problem comes from my version C exam.
65
00:06:46.000 --> 00:06:51.520
That's right. Now, the version C exam, that's the exam I give to the student who like,
66
00:06:51.520 --> 00:06:58.720
like to miss class a lot, or want to make up a test. What does C stand for? Oh, what does the
67
00:06:58.720 --> 00:07:05.840
C stand for? The C stands for CU X and Nester. Anyway, you don't want the version C exam, so come
68
00:07:05.840 --> 00:07:10.320
to class every day. Anyway, let's go ahead and finish this problem, because we're not done yet.
69
00:07:11.760 --> 00:07:19.920
Okay, what do we have to do, Charlie? We have to replace the U with tangent of X over 2. So,
70
00:07:19.920 --> 00:07:27.200
here we go. Okay, well, first, we can combine those logarithms, right? Because notice, it's 1 fifth
71
00:07:27.200 --> 00:07:33.200
natural log of absolute value of 3 minus 1, subtract 1 fifth natural log of U plus 3 plus C. And
72
00:07:33.200 --> 00:07:38.480
remember, the difference of the logs is the log of the quotient. So, we can write this as 1 fifth
73
00:07:38.480 --> 00:07:44.560
natural log of what, Charlie? 3 minus 1 over U plus 3, that's right, plus C. And now, we can finally
74
00:07:44.560 --> 00:07:49.840
replace the U with what, Charlie? Tangent of X over 2. Tangent of X over 2. So, what do we get on top?
75
00:07:49.840 --> 00:07:53.600
3 tangent of X over 2 minus 1. That's right. And what do we get in the bottom? Tangent of X over 2 plus
76
00:07:53.600 --> 00:08:00.720
3. And then don't forget to put C. Now, Charlie, we're still not done with this problem. There's still
77
00:08:00.720 --> 00:08:08.080
one more thing to do. What is it? What? Circle your answer. Very nice. There. Because that was a
78
00:08:08.080 --> 00:08:14.160
good warm-up problem. Anyway, we're going to come back later and do some more problems. So anyway,
79
00:08:14.160 --> 00:08:44.000
we'll see you all again soon.