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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 you write any such

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expression as a sum of things,
each of which is relatively

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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 that we've

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already learned.

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So I have here 4 rational
functions for you.

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And what I'd like you to do in
each case, is try to decompose

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it into the general form that
Professor Jerison taught you.

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So don't, I'm not asking you
to, if you'd like, you're

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certainly welcome to go
ahead and compute the

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antiderivatives after you do
that, but I'm not going to do

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it for you, or I'm not going
to ask you to do it.

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So what I'd like you to do,
though, is for each of these

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four expressions, to break it
out into the form of the

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partial fraction
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, or I've picked them to

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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 over

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x squared minus 8x.

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Now, the first thing to do when
you start the partial

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fraction method, is that you
have to check that the degree

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of the numerator is smaller
than the degree of the

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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--

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well, 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, and it's not too hard to

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check that this is equal to
1 plus 4x plus 4 over x

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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 get

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something in front that's
always a polynomial.

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And that's good, because
remember, our goal is to, you

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know, manipulate this into a
form where we can integrate

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it, and polynomials 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 decompose the

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second part.

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So to do that, the first thing
you need to do, once you've

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got a smaller degree on top than
downstairs, is that you

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factor the denominator.

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So in this case, I'm going to
keep the 1 plus, just so I can

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keep writing equals signs
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, and we see that

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the denominator is x
times x-- whoops--

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minus 8.

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OK.

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Now, in this case, this is the
simplest situation for partial

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fraction decomposition.

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We have the denominator
is a product of

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distinct linear factors.

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So when the denominator is a
product of distinct linear

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factors, what we get this is
equal to, what the partial

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fraction theorem tells us we
can write, is that this is

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equal to something, some
constant, over the first

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factor plus some constant
over the second factor.

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Now, if you had, you know, three
factors, then you'd have

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three of these terms, one for
each factor, if they were

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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 that Professor

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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, and we want to find

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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

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true, 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, and this is going

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to give us a logarithm.

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So once we find these constants
a and b, we're all

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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, everything that doesn't

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x in the denominator,
and also x.

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And then you go back here, and
you plug in the appropriate

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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, which

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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 everything that

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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, and on

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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, it's 36, and

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divided by eight, so that's
36 divided by 8 is

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9 divided by 2.

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So once you've got these
values of a and b--

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OK.

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So our original expression,
now we substitute in these

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values of a and b, then we know
that this is is a true

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equality, and then the
integration of this expression

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is just reduced completely to
the integration of this

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easy-to-integrate expression.

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OK?

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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 9/2.

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So that's part A. Let's go on
to part B. I have to remind

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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, we still have

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only linear factors, 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

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fraction decomposition looks
a little bit different.

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And so what you get is that you
have, for every repeated

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power, so for each--

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this one appears four times, so
you get 4 summands on the

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right hand side, with increasing
powers of this

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linear factor.

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So with increasing powers
of x minus 1.

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So this is going to be a--

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or sorry, x plus 1.

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a over x plus 1 plus b over x
plus 1 squared plus c over x

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plus 1 cubed plus d over
x plus 1 to the fourth.

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Now remember, even though the
degree here goes up in these

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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 got the right--

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you know, before you solve for
the constants, make sure

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you've got the correct abstract
decomposition--

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is to count the number of these
constants that you're

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looking for.

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It should always match
the degree of the

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denominator over here.

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So in this case, the degree of
the denominator is 4, and

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there are 4 constants.

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So we've, so that's a good way
to check that you set the

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problem up right.

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OK.

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So now, what do we
do in this case?

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Well, the cover up method works,
but it only works to

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find the highest degree term.

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Right?

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So we cover up x plus 1 to the
fourth, and we cover up

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everything with a smaller power
of x plus 1, and then we

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plug in negative 1.

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Right?

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Because we need x
plus 1 to be 0.

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So OK, so over here we get
negative 1 squared is 1, so we

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get, right away, that
d is equal to 1.

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OK.

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But that doesn't give
us a, b, or c.

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We can't get a, b, or c by
the cover up method.

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Now, there are a couple
different ways you can proceed

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at this point.

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One thing you can do, which is
something that often works, is

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you could plug in values.

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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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And as you plug in other values
for x, what you'll see

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is that for every value you plug
in, you'll get a linear

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equation relating your
variables, a, b, c, and

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whatever, in this case, that's
all we've got left, a, b, and

203
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c, to each other.

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And so if you plug in three
different values of x, say,

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you'll get three different
linear equations with three

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different variables, and then
you can solve them.

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That's one thing you can do.

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Another thing you could do is
you can multiply through by x

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plus 1 to the fourth.

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So if you do that, you'll have
on the left, you'll just have

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00:10:25,610 --> 00:10:28,720
x squared, and on the right
you'll have-- well, you'll

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have a times x plus 1 cubed plus
b times x plus 1 squared,

213
00:10:33,940 --> 00:10:36,340
plus c times x plus 1, plus d.

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And we already know that
x is equal to 1.

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OK?

216
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And so then, for those two
things to be equal, they're

217
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equal as polynomials, all their

218
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coefficients have to be equal.

219
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So you can just look at the
coefficients in that resulting

220
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expression, and ask, you know,
which coefficients--

221
00:10:55,680 --> 00:10:56,500
sorry.

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You can set coefficients
on the two sides equal.

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The two polynomials are equal,
all of their corresponding

224
00:11:01,120 --> 00:11:02,090
coefficients are equal.

225
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So you could look, you know,
at this side and

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you'll say, oh, OK.

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00:11:04,570 --> 00:11:08,090
So the coefficient of x cubed
has to be the same as whatever

228
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the coefficient of x cubed
over here is, and the

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coefficient x squared has to be
the same as the coefficient

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of x squared, and so on.

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So that's another
way to proceed.

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00:11:17,010 --> 00:11:24,690

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Yeah, all right.

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Those are really your
two best options.

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I like to multiply through,
personally.

236
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So OK.

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So if I were to do that, in this
case, on the left hand

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side I'd x squared equals a
times x plus 1 cubed plus b

239
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times x plus 1 squared plus
c times x plus 1 plus d.

240
00:11:54,470 --> 00:11:56,860
Except we already know that d
is equal to 1, so I'm just

241
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going to write plus 1.

242
00:11:58,670 --> 00:12:01,700

243
00:12:01,700 --> 00:12:02,230
So OK.

244
00:12:02,230 --> 00:12:04,750
So actually, let me say
that there are other

245
00:12:04,750 --> 00:12:05,610
things you could do.

246
00:12:05,610 --> 00:12:08,890
Which is, you could rearrange
things and simplify

247
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algebraically before plugging in
values, or before comparing

248
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coefficients.

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So let me give you one example
of each of those three

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00:12:19,010 --> 00:12:19,800
possibilities.

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00:12:19,800 --> 00:12:23,600
So for example, one thing you
can do, is you can look at the

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00:12:23,600 --> 00:12:25,430
highest degree coefficient.

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00:12:25,430 --> 00:12:27,830
So as Professor Jerison said,
the easiest coefficients are

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00:12:27,830 --> 00:12:31,210
usually the high order terms and
the low order terms. So in

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00:12:31,210 --> 00:12:33,110
this case, the high order
terms would be--

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00:12:33,110 --> 00:12:35,760
this is a third degree
polynomial on the right hand

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00:12:35,760 --> 00:12:38,180
side, and it's a second degree
polynomial on the left.

258
00:12:38,180 --> 00:12:41,130
So the highest order term here,
x cubed, just appears in

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00:12:41,130 --> 00:12:43,520
this one place as
coefficient a.

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00:12:43,520 --> 00:12:43,700
Right?

261
00:12:43,700 --> 00:12:47,670
This is ax cubed plus something
times x squared plus

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00:12:47,670 --> 00:12:48,630
blah blah blah.

263
00:12:48,630 --> 00:12:51,640
And over here, we have
no x cubeds.

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00:12:51,640 --> 00:12:54,140
So we have x cubeds here,
but no x cubeds here.

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00:12:54,140 --> 00:12:56,500
That means the coefficient of x
cubed here has to be 0, so a

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00:12:56,500 --> 00:12:57,750
has to be 0.

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00:12:57,750 --> 00:13:01,160

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

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00:13:02,050 --> 00:13:05,270
So a has to be equal to 0, and
that simplifies everything a

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00:13:05,270 --> 00:13:05,630
little bit.

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00:13:05,630 --> 00:13:11,960
So now we get x squared equals
b times x plus 1 squared plus

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00:13:11,960 --> 00:13:18,310
c times x plus 1 plus 1.

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00:13:18,310 --> 00:13:22,240
Now let me show you what
I mean about algebraic

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00:13:22,240 --> 00:13:23,630
manipulation.

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00:13:23,630 --> 00:13:26,490
This one, if you wanted, you
could always just subtract it

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00:13:26,490 --> 00:13:27,660
over to the other side.

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

278
00:13:28,190 --> 00:13:30,060
And so then you'll have, on
the left you'll have x

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00:13:30,060 --> 00:13:31,390
squared minus 1.

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00:13:31,390 --> 00:13:35,910
And x squared minus 1, you can
write as x minus 1 times x

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00:13:35,910 --> 00:13:46,770
plus 1 equals b times x plus 1
squared plus c times x plus 1.

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00:13:46,770 --> 00:13:48,795
And then you can divide out
by an x plus 1 everywhere.

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00:13:48,795 --> 00:13:54,600
It appears in all terms. So
you get x minus 1 equals b

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00:13:54,600 --> 00:13:59,570
times x plus 1 plus c.

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00:13:59,570 --> 00:14:02,080
And now what this does for
you, is you sort of just

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00:14:02,080 --> 00:14:03,400
reduced the degree everywhere.

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

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

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00:14:09,960 --> 00:14:12,590
And here, so for example, if you
substitute x equals minus

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00:14:12,590 --> 00:14:16,790
1, that's the same idea
as what you do in

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00:14:16,790 --> 00:14:18,330
the cover up method.

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00:14:18,330 --> 00:14:21,370
This b term will just die
completely, and you'll be left

293
00:14:21,370 --> 00:14:22,860
with negative 2 on the left.

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00:14:22,860 --> 00:14:24,460
So you get c--

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00:14:24,460 --> 00:14:26,780
I'm going to have to move
over here, sorry.

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

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00:14:28,540 --> 00:14:32,040
And also, you can do the one
thing that I haven't done so

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00:14:32,040 --> 00:14:35,130
far, is this plugging in
nice choices of values.

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00:14:35,130 --> 00:14:37,570
So another nice choice of value
for x that we haven't

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00:14:37,570 --> 00:14:39,370
used is x equals 0.

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

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00:14:42,230 --> 00:14:44,001
equals b plus c.

303
00:14:44,001 --> 00:14:45,251
So minus 1 equals b plus c.

304
00:14:45,251 --> 00:14:48,610

305
00:14:48,610 --> 00:14:50,960
And since we just found
c, that means that

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00:14:50,960 --> 00:14:54,830
b is equal to 1.

307
00:14:54,830 --> 00:14:55,260
All right.

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00:14:55,260 --> 00:14:56,990
So--

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00:14:56,990 --> 00:14:59,200
oh boy, I'm using a lot of
space, aren't I. All right.

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00:14:59,200 --> 00:15:02,580
So in this case, we've got our
coefficients, a equals 0, d

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00:15:02,580 --> 00:15:05,580
equals 1, b equals 1,
c equals minus 2.

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

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00:15:09,400 --> 00:15:16,190
Let's go back over here then,
and look at question C. So for

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00:15:16,190 --> 00:15:19,080
C, the question is, what
is the partial fraction

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00:15:19,080 --> 00:15:23,840
decomposition of 2x plus 2
divided by the quantity 4x

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00:15:23,840 --> 00:15:27,710
squared plus 1 squared?

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00:15:27,710 --> 00:15:28,450
This one's really easy.

318
00:15:28,450 --> 00:15:29,880
This one is done.

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00:15:29,880 --> 00:15:32,320
This is already partial
fraction decomposed.

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00:15:32,320 --> 00:15:32,690
Right?

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

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

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

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

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

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00:15:47,680 --> 00:15:52,650
you want something linear over
4x squared plus 1, plus

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00:15:52,650 --> 00:15:56,760
something linear over 4x
squared plus 1 squared.

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

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00:15:58,690 --> 00:16:02,970
The first linear part is 0, and
the second is something

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00:16:02,970 --> 00:16:05,490
linear over 4x squared
plus 1 squared.

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00:16:05,490 --> 00:16:08,540
So to integrate this, it's
already in a pretty good form.

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00:16:08,540 --> 00:16:10,060
Now, you're actually going to
write, if you wanted to

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00:16:10,060 --> 00:16:13,540
integrate this, you would split
it into two pieces, one

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00:16:13,540 --> 00:16:15,900
with the 2x and then
one with the 2.

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00:16:15,900 --> 00:16:19,080
And the first one, you would
just be a usual u

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00:16:19,080 --> 00:16:21,660
substitution, and the second
one, we would want some sort

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00:16:21,660 --> 00:16:24,190
of trigonometric substitution.

338
00:16:24,190 --> 00:16:28,450
But this one is already ready
for methods we already should

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00:16:28,450 --> 00:16:29,670
be comfortable with.

340
00:16:29,670 --> 00:16:29,930
OK.

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

342
00:16:31,090 --> 00:16:32,255
It's done already.

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

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

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00:16:36,160 --> 00:16:40,790
And for, all right, and so for
this last one, I'm also not

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00:16:40,790 --> 00:16:42,470
going to write this one out.

347
00:16:42,470 --> 00:16:45,250
But the thing to notice here
is the way I wrote it-- and

348
00:16:45,250 --> 00:16:46,350
this was really mean of me.

349
00:16:46,350 --> 00:16:46,890
Right?

350
00:16:46,890 --> 00:16:49,630
I wrote it as x squared minus
1 quantity squared.

351
00:16:49,630 --> 00:16:52,240
So a natural instinct
is to say, aha!

352
00:16:52,240 --> 00:16:55,090
It's a quadratic repeated factor
in the denominator.

353
00:16:55,090 --> 00:16:55,340
Right?

354
00:16:55,340 --> 00:16:58,490
But that's just because I was
mean and I wrote it this way.

355
00:16:58,490 --> 00:16:59,870
That's not actually
what this is.

356
00:16:59,870 --> 00:17:01,810
This is not irreducible.

357
00:17:01,810 --> 00:17:02,760
This factors.

358
00:17:02,760 --> 00:17:04,850
You can rewrite--

359
00:17:04,850 --> 00:17:07,723
let me come back over here.

360
00:17:07,723 --> 00:17:15,500
This is for question D. So you
can rewrite x squared minus 1

361
00:17:15,500 --> 00:17:22,100
squared as x minus 1 squared
times x plus 1 squared.

362
00:17:22,100 --> 00:17:27,190
You can factor this
x squared minus 1.

363
00:17:27,190 --> 00:17:29,550
So when you factor this x minus
1, what you see is--

364
00:17:29,550 --> 00:17:33,900
this isn't a problem that has
one irreducible quadratic

365
00:17:33,900 --> 00:17:36,230
factor appearing to
the second power.

366
00:17:36,230 --> 00:17:39,590
What it has is two linear
factors, each appearing to the

367
00:17:39,590 --> 00:17:40,840
second power.

368
00:17:40,840 --> 00:17:43,700
So the partial fraction
decomposition in this problem

369
00:17:43,700 --> 00:17:52,890
will be something like a over x
minus 1 plus b over x minus

370
00:17:52,890 --> 00:17:59,230
1 squared plus c over
x plus 1, plus d

371
00:17:59,230 --> 00:18:03,180
over x plus 1 squared.

372
00:18:03,180 --> 00:18:06,110
That's what you'll get when
you apply partial fraction

373
00:18:06,110 --> 00:18:07,450
decomposition to this problem.

374
00:18:07,450 --> 00:18:10,660
And then you have to solve
for the coefficients

375
00:18:10,660 --> 00:18:11,530
a, b, c, and d.

376
00:18:11,530 --> 00:18:15,650
So I'm not going to write that
out myself, but I cleverly did

377
00:18:15,650 --> 00:18:19,630
it before I came on camera, so I
can tell you what the answer

378
00:18:19,630 --> 00:18:21,310
is, if you want to
check your work.

379
00:18:21,310 --> 00:18:32,290
So here we have a is equal to 0,
b is equal to 1, c is equal

380
00:18:32,290 --> 00:18:35,690
to 1, and d is equal
to minus 3.

381
00:18:35,690 --> 00:18:38,300
So that's for the, I didn't
write it over here.

382
00:18:38,300 --> 00:18:41,940
That's for this particular
numerator that we started with

383
00:18:41,940 --> 00:18:43,950
back over here.

384
00:18:43,950 --> 00:18:46,780
So for this particular fraction,
if you carry out the

385
00:18:46,780 --> 00:18:51,610
partial fraction decomposition,
what you get is

386
00:18:51,610 --> 00:18:53,810
right here.

387
00:18:53,810 --> 00:18:56,420
So OK, so those are, that was
a few more examples of the

388
00:18:56,420 --> 00:18:58,210
partial fraction
decomposition.

389
00:18:58,210 --> 00:19:02,560
I hope you enjoyed them, and
I'm going to end there.

390
00:19:02,560 --> 00:19:02,609