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Hi.
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Welcome back to recitation.
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You've been talking in
class about partial fraction
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decomposition as a
tool for integration.
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So remember that the point of
partial fraction decomposition
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is that whenever you have a
rational function, that is,
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one polynomial divided by
another, that in principle,
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partial fraction
decomposition lets
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you write any such expression as
a sum of things, each of which
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is relatively easy to integrate.
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So the technique here
is purely algebraic.
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And then you just
apply integral rules
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that we've already learned.
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So I have here four
rational functions for you.
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And what I'd like you
to do in each case,
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is try to decompose it
into the general form
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that Professor
Jerison taught you.
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So don't, I'm not asking
you to-- if you'd like,
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you're certainly welcome
to go ahead and compute
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the antiderivatives after
you do that, but I'm
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not going to do it
for you, or I'm not
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going to ask you to do it.
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So what I'd like
you to do, though,
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is for each of these
four expressions,
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to break it out into the
form of the partial fraction
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decomposition.
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So why don't you take a few
minutes to do that, come back,
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and you can check your
work against mine.
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All right.
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Welcome back.
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Hopefully you had some
fun working on these.
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They're a little
bit tricky, I think,
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or I've picked them to
be a little bit tricky.
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So let's go through
them one by one.
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I guess I'll start
with the first one.
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So with the first one, I have
x squared minus 4x plus 4
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over x squared minus 8x.
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Now, the first thing to do when
you start the partial fraction
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method, is that
you have to check
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that the degree of
the numerator is
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smaller than the degree
of the denominator.
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Now in this case,
that's not true.
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The degree on top is 2, and
the degree on bottom is 2.
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So we need to do long division.
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Or you know, we need
to do something-- well,
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long division is the usual
and always-- usual way,
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and always works-- in order to
reduce the degree of the top
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here so that it's smaller
than the degree at the bottom.
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Now, I've done
this ahead of time,
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and it's not too hard
to check that this
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is equal to 1 plus 4x plus
4 over x squared minus 8x.
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It's a relatively easy long
division to do in any case.
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So OK.
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So what we get after we
do that process is we
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get something in front
that's always a polynomial.
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And that's good,
because remember,
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our goal is to, you know,
manipulate this into a form
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where we can integrate
it, and polynomials
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are easy to integrate.
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So then we usually just forget
about this for the time being.
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And what we want to
do is partial fraction
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decompose the second part.
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So to do that, the first
thing you need to do,
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once you've got a smaller
degree on top than downstairs,
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is that you factor
the denominator.
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So in this case, I'm
going to keep the 1 plus,
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just so I can keep
writing equals signs
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and be honest about it.
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But really, our focus
now is just entirely
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on this second summand.
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So this is equal to 1
plus 4x plus 4 is on top.
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And OK, so we need to factor
the denominator, if possible.
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And in this case, that's
not only possible.
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It's pretty straightforward.
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We can factor out an
x from both terms,
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and we see that
the denominator is
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x times x-- whoops-- minus 8.
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OK.
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Now, in this case, this
is the simplest situation
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for partial fraction
decomposition.
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We have the denominator
is a product
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of distinct linear factors.
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So when the denominator
is a product
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of distinct linear factors,
what we get this is equal to,
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what the partial fraction
theorem tells us we can write,
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is that this is equal to
something, some constant,
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over the first factor
plus some constant
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over the second factor.
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Now, if you had, you know,
three factors, then you'd
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have three of these terms,
one for each factor,
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if they were distinct
linear factors.
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So great.
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OK.
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Now we can apply
the cover up method
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that Professor
Jerison taught us.
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So again, the 1 doesn't
really matter here.
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You can just ignore it.
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So we have this is
equal to this sum,
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and we want to find the values
of A and B that make this true.
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And then once we have values
of A and B that make this true,
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the resulting expression
will be all set to integrate.
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Right?
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This will just be easy.
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It's just a polynomial,
in fact, just a constant.
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This is going to
give us a logarithm,
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and this is going to
give us a logarithm.
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So once we find these
constants A and B,
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we're all set to integrate.
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So OK, so how does the
cover up method work?
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Well, so you cover up
one of the factors.
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In this case, x.
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And you cover up,
on the other side,
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everything that doesn't have x
in the denominator, and also x.
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And then you go
back here, and you
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plug in the appropriate values,
you know, the x minus whatever.
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So in this case,
that's x equals 0.
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And so over here, you
get 4 over minus 8.
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So that gives us A is equal
to 4 divided by negative 8,
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which is minus a half.
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And now we can do the same
thing with the x minus 8 part.
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So we cover up x
minus 8, we cover up
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everything that doesn't
have an x minus 8 in it.
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So we've got, on the
right-hand side, we get B,
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and on the left hand side,
we have to plug in 8 here.
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So we plug 4 times 8 plus 4,
which is 32 plus-- it's 36,
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and divided by eight, so
that's 36 divided by 8
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is 9 divided by 2.
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So once you've got these
values of A and B-- OK.
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So our original
expression, now we
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substitute in these
values of A and B,
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then we know that this
is is a true equality,
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and then the integration
of this expression
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is just reduced completely
to the integration
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of this easy-to-integrate
expression.
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OK?
136
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And so this is the partial
fraction decomposition here,
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with, you know A equals minus a
half, and B equals nine halves.
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So that's part (a).
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Let's go on to part (b).
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I have to remind myself
what part (b) is.
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OK.
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So part (b) is x squared divided
by x plus 2 to the fourth.
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So in this case,
in the denominator,
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we still have only
linear factors,
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but we have repeated factors.
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And I mean, this is actually
a particularly simple example,
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where we have only one factor,
but it's repeated four times.
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Right?
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Fourth power.
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So when you have that
situation, the partial fraction
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decomposition looks a
little bit different.
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And so what you get is that you
have, for every repeated power,
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so for each-- this one
appears four times,
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so you get four summands
on the right-hand side,
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with increasing powers
of this linear factor.
156
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So with increasing
powers of x minus 1.
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So this is going to be
A-- or sorry, x plus 1.
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A over x plus 1
plus B over x plus 1
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squared plus C over x plus
1 cubed plus D over x plus 1
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to the fourth.
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Now remember, even
though the degree here
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goes up in these later summands,
what stays on top is the same.
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It just stays a constant.
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If this were a quadratic
factor, it would stay linear.
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The top doesn't increase
in degree when do this.
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And a good simple way
to check that you've
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got the right-- you know, before
you solve for the constants,
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make sure you've got the
correct abstract decomposition--
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is to count the number
of these constants
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that you're looking for.
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It should always
match the degree
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of the denominator over here.
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So in this case, the degree
of the denominator is 4,
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and there are 4 constants.
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So we've-- so that's a good
way to check that you set
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the problem up right.
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OK.
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00:08:54,940 --> 00:08:57,020
So now, what do we
do in this case?
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Well, the cover up
method works, but it only
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works to find the
highest degree term.
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Right?
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00:09:03,090 --> 00:09:06,830
So we cover up x
plus 1 to the fourth,
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00:09:06,830 --> 00:09:11,940
and we cover up everything with
a smaller power of x plus 1,
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00:09:11,940 --> 00:09:14,880
and then we plug in negative 1.
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Right?
186
00:09:15,380 --> 00:09:17,730
Because we need
x plus 1 to be 0.
187
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So OK, so over here we get
negative 1 squared is 1,
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so we get, right away,
that D is equal to 1.
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OK.
190
00:09:28,010 --> 00:09:30,460
But that doesn't
give us A, B, or C.
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00:09:30,460 --> 00:09:34,860
We can't get A, B, or C
by the cover up method.
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00:09:34,860 --> 00:09:36,510
Now, there are a
couple different ways
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00:09:36,510 --> 00:09:38,730
you can proceed at this point.
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00:09:38,730 --> 00:09:43,910
One thing you can do, which
is something that often works,
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is you could plug in values.
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00:09:47,230 --> 00:09:48,660
Well, so this will always work.
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I shouldn't say often works.
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You can start plugging
in other values for x.
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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.
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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.