Hi.
Welcome back to recitation.
You've been talking in
class about partial fraction
decomposition as a
tool for integration.
So remember that the point of
partial fraction decomposition
is that whenever you have a
rational function, that is,
one polynomial divided by
another, that in principle,
partial fraction
decomposition lets
you write any such expression as
a sum of things, each of which
is relatively easy to integrate.
So the technique here
is purely algebraic.
And then you just
apply integral rules
that we've already learned.
So I have here four
rational functions for you.
And what I'd like you
to do in each case,
is try to decompose it
into the general form
that Professor
Jerison taught you.
So don't, I'm not asking
you to-- if you'd like,
you're certainly welcome
to go ahead and compute
the antiderivatives after
you do that, but I'm
not going to do it
for you, or I'm not
going to ask you to do it.
So what I'd like
you to do, though,
is for each of these
four expressions,
to break it out into the
form of the partial fraction
decomposition.
So why don't you take a few
minutes to do that, come back,
and you can check your
work against mine.
All right.
Welcome back.
Hopefully you had some
fun working on these.
They're a little
bit tricky, I think,
or I've picked them to
be a little bit tricky.
So let's go through
them one by one.
I guess I'll start
with the first one.
So with the first one, I have
x squared minus 4x plus 4
over x squared minus 8x.
Now, the first thing to do when
you start the partial fraction
method, is that
you have to check
that the degree of
the numerator is
smaller than the degree
of the denominator.
Now in this case,
that's not true.
The degree on top is 2, and
the degree on bottom is 2.
So we need to do long division.
Or you know, we need
to do something-- well,
long division is the usual
and always-- usual way,
and always works-- in order to
reduce the degree of the top
here so that it's smaller
than the degree at the bottom.
Now, I've done
this ahead of time,
and it's not too hard
to check that this
is equal to 1 plus 4x plus
4 over x squared minus 8x.
It's a relatively easy long
division to do in any case.
So OK.
So what we get after we
do that process is we
get something in front
that's always a polynomial.
And that's good,
because remember,
our goal is to, you know,
manipulate this into a form
where we can integrate
it, and polynomials
are easy to integrate.
So then we usually just forget
about this for the time being.
And what we want to
do is partial fraction
decompose the second part.
So to do that, the first
thing you need to do,
once you've got a smaller
degree on top than downstairs,
is that you factor
the denominator.
So in this case, I'm
going to keep the 1 plus,
just so I can keep
writing equals signs
and be honest about it.
But really, our focus
now is just entirely
on this second summand.
So this is equal to 1
plus 4x plus 4 is on top.
And OK, so we need to factor
the denominator, if possible.
And in this case, that's
not only possible.
It's pretty straightforward.
We can factor out an
x from both terms,
and we see that
the denominator is
x times x-- whoops-- minus 8.
OK.
Now, in this case, this
is the simplest situation
for partial fraction
decomposition.
We have the denominator
is a product
of distinct linear factors.
So when the denominator
is a product
of distinct linear factors,
what we get this is equal to,
what the partial fraction
theorem tells us we can write,
is that this is equal to
something, some constant,
over the first factor
plus some constant
over the second factor.
Now, if you had, you know,
three factors, then you'd
have three of these terms,
one for each factor,
if they were distinct
linear factors.
So great.
OK.
Now we can apply
the cover up method
that Professor
Jerison taught us.
So again, the 1 doesn't
really matter here.
You can just ignore it.
So we have this is
equal to this sum,
and we want to find the values
of A and B that make this true.
And then once we have values
of A and B that make this true,
the resulting expression
will be all set to integrate.
Right?
This will just be easy.
It's just a polynomial,
in fact, just a constant.
This is going to
give us a logarithm,
and this is going to
give us a logarithm.
So once we find these
constants A and B,
we're all set to integrate.
So OK, so how does the
cover up method work?
Well, so you cover up
one of the factors.
In this case, x.
And you cover up,
on the other side,
everything that doesn't have x
in the denominator, and also x.
And then you go
back here, and you
plug in the appropriate values,
you know, the x minus whatever.
So in this case,
that's x equals 0.
And so over here, you
get 4 over minus 8.
So that gives us A is equal
to 4 divided by negative 8,
which is minus a half.
And now we can do the same
thing with the x minus 8 part.
So we cover up x
minus 8, we cover up
everything that doesn't
have an x minus 8 in it.
So we've got, on the
right-hand side, we get B,
and on the left hand side,
we have to plug in 8 here.
So we plug 4 times 8 plus 4,
which is 32 plus-- it's 36,
and divided by eight, so
that's 36 divided by 8
is 9 divided by 2.
So once you've got these
values of A and B-- OK.
So our original
expression, now we
substitute in these
values of A and B,
then we know that this
is is a true equality,
and then the integration
of this expression
is just reduced completely
to the integration
of this easy-to-integrate
expression.
OK?
And so this is the partial
fraction decomposition here,
with, you know A equals minus a
half, and B equals nine halves.
So that's part (a).
Let's go on to part (b).
I have to remind myself
what part (b) is.
OK.
So part (b) is x squared divided
by x plus 2 to the fourth.
So in this case,
in the denominator,
we still have only
linear factors,
but we have repeated factors.
And I mean, this is actually
a particularly simple example,
where we have only one factor,
but it's repeated four times.
Right?
Fourth power.
So when you have that
situation, the partial fraction
decomposition looks a
little bit different.
And so what you get is that you
have, for every repeated power,
so for each-- this one
appears four times,
so you get four summands
on the right-hand side,
with increasing powers
of this linear factor.
So with increasing
powers of x minus 1.
So this is going to be
A-- or sorry, x plus 1.
A over x plus 1
plus B over x plus 1
squared plus C over x plus
1 cubed plus D over x plus 1
to the fourth.
Now remember, even
though the degree here
goes up in these later summands,
what stays on top is the same.
It just stays a constant.
If this were a quadratic
factor, it would stay linear.
The top doesn't increase
in degree when do this.
And a good simple way
to check that you've
got the right-- you know, before
you solve for the constants,
make sure you've got the
correct abstract decomposition--
is to count the number
of these constants
that you're looking for.
It should always
match the degree
of the denominator over here.
So in this case, the degree
of the denominator is 4,
and there are 4 constants.
So we've-- so that's a good
way to check that you set
the problem up right.
OK.
So now, what do we
do in this case?
Well, the cover up
method works, but it only
works to find the
highest degree term.
Right?
So we cover up x
plus 1 to the fourth,
and we cover up everything with
a smaller power of x plus 1,
and then we plug in negative 1.
Right?
Because we need
x plus 1 to be 0.
So OK, so over here we get
negative 1 squared is 1,
so we get, right away,
that D is equal to 1.
OK.
But that doesn't
give us A, B, or C.
We can't get A, B, or C
by the cover up method.
Now, there are a
couple different ways
you can proceed at this point.
One thing you can do, which
is something that often works,
is you could plug in values.
Well, so this will always work.
I shouldn't say often works.
You can start plugging
in other values for x.
And as you plug in other
values for x, what you'll see
is that for every value
you plug in, you'll
get a linear equation
relating your variables-- A,
B, C, and whatever,
in this case, that's
all we've got left, A,
B, and C-- to each other.
And so if you plug in three
different values of x, say,
you'll get three
different linear equations
with three different variables,
and then you can solve them.
That's one thing you can do.
Another thing you could do is
you can multiply through by x
plus 1 to the fourth.
So if you do that, you'll
have-- on the left,
you'll just have x
squared, and on the right
you'll have-- well,
you'll have A times
x plus 1 cubed plus B times x
plus 1 squared, plus C times
x plus 1, plus D. And we already
know that x is equal to 1.
OK?
And so then, for those
two things to be equal,
they're equal as
polynomials, all
their coefficients
have to be equal.
So you can just look at the
coefficients in that resulting
expression, and ask, you know,
which coefficients-- sorry.
You can set coefficients
on the two sides equal.
The two polynomials
are equal, all
of their corresponding
coefficients are equal.
So you could look,
you know, at this side
and you'll say, oh, OK.
So the coefficient
of x cubed has
to be the same as whatever the
coefficient of x cubed over
here is, and the
coefficient x squared
has to be the same as the
coefficient of x squared,
and so on.
So that's another
way to proceed.
Yeah, all right.
Those are really your
two best options.
I like to multiply
through, personally.
So OK.
So if I were to do that, in
this case, on the left-hand side
I'd get x squared equals
A times x plus 1 cubed
plus B times x plus 1 squared
plus C times x plus 1 plus D.
Except we already know
that D is equal to 1,
so I'm just going
to write plus 1.
So OK.
So actually, let
me say that there
are other things you could do.
Which is, you could
rearrange things
and simplify algebraically
before plugging in values,
or before comparing
coefficients.
So let me give you
one example of each
of those three possibilities.
So for example, one
thing you can do,
is you can look at the
highest degree coefficient.
So as Professor Jerison said,
the easiest coefficients
are usually the high-order
terms and the low-order terms.
So in this case,
the high-order terms
would be-- this
is a third degree
polynomial on the
right-hand side,
and it's a second degree
polynomial on the left.
So the highest-order
term here, x cubed,
just appears in this one
place as coefficient A. Right?
This is A x cubed
plus something times x
squared plus blah blah blah.
And over here, we
have no x cubeds.
So we have x cubeds here,
but no x cubeds here.
That means the coefficient
of x cubed here has to be 0,
so A has to be 0.
OK.
So A has to be equal to 0,
and that simplifies everything
a little bit.
So now we get x squared equals
B times x plus 1 squared
plus C times x plus 1 plus 1.
Now let me show you what I mean
about algebraic manipulation.
This 1, if you wanted, you could
always just subtract it over
to the other side.
Right?
And so then you'll
have, on the left
you'll have x squared minus 1.
And x squared minus
1, you can write
as x minus 1 times x plus 1
equals B times x plus 1 squared
plus C times x plus 1.
And then you can divide out
by an x plus 1 everywhere.
It appears in all terms.
So you get x minus 1 equals
B times x plus 1 plus C.
And now what this does for you,
is you sort of just reduced
the degree everywhere.
And actually, you could
substitute x equals minus 1
again, if you wanted
to, for example.
And here, so for example, if
you substitute x equals minus 1,
that's the same idea as what
you do in the cover up method.
This B term will
just die completely,
and you'll be left with
negative 2 on the left.
So you get C-- I'm going to
have to move over here, sorry.
C is equal to negative 2, then.
And also, you can
do the one thing
that I haven't done so
far, is this plugging
in nice choices of values.
So another nice choice of value
for x that we haven't used
is x equals 0.
So if you plug in x equals
0, you'll get minus 1
equals B plus c.
So minus 1 equals B plus C.
And since we just found C,
that means that B is equal to 1.
All right.
So-- oh boy, I'm using a lot
of space, aren't I. All right.
So in this case, we've got
our coefficients, A equals 0,
D equals 1, B equals
1, C equals minus 2.
And that gives us the partial
fraction decomposition.
Let's go back over here then,
and look at question (c).
So for (c), the question is,
what is the partial fraction
decomposition of 2x plus
2 divided by the quantity
4x squared plus 1 squared?
This one's really easy.
This one is done.
This is already partial
fraction decomposed.
Right?
When you have-- so here we
have an irreducible quadratic
in the denominator.
You can't factor this any
further than it's gone.
It also occurs to
a higher power.
So when you partial fraction
decompose something like this,
you want something linear
over 4x squared plus 1,
plus something linear over
4x squared plus 1 squared.
But we already have that, right?
The first linear part
is 0, and the second
is something linear over
4x squared plus 1 squared.
So to integrate this, it's
already in a pretty good form.
Now, you're actually
going to write,
if you wanted to
integrate this, you
would split it into two
pieces, one with the 2x
and then one with the 2.
And the first one,
you would just
be a usual u substitution,
and the second one,
we would want some sort of
trigonometric substitution.
But this one is already
ready for methods we already
should be comfortable with.
OK.
So C, that's easy.
It's done already.
I'll put a check mark there,
because that makes me happy.
OK.
And for-- all right, and
so for this last one,
I'm also not going to
write this one out.
But the thing to notice
here is the way I wrote it--
and this was really mean of me.
Right?
I wrote it as x squared
minus 1 quantity squared.
So a natural instinct
is to say, aha!
It's a quadratic repeated
factor in the denominator.
Right?
But that's just because I was
mean and I wrote it this way.
That's not actually
what this is.
This is not irreducible.
This factors.
You can rewrite-- let
me come back over here.
This is for question (d).
So you can rewrite x squared
minus 1 squared as x minus 1
squared times x plus 1 squared.
You can factor this
x squared minus 1.
So when you factor this
x minus 1, what you see
is-- this isn't a
problem that has
one irreducible quadratic factor
appearing to the second power.
What it has is two
linear factors,
each appearing to
the second power.
So the partial fraction
decomposition in this problem
will be something
like A over x minus 1
plus B over x minus 1
squared plus C over x plus 1,
plus D over x plus 1 squared.
That's what you'll get when
you apply partial fraction
decomposition to this problem.
And then you have to solve
for the coefficients A, B, C,
and D.
So I'm not going to write that
out myself, but I cleverly
did it before I
came on camera, so I
can tell you what the answer is,
if you want to check your work.
So here we have A is equal
to 0, B is equal to 1,
C is equal to 1, and
D is equal to minus 3.
So that's for the-- I
didn't write it over here.
That's for this
particular numerator
that we started
with back over here.
So for this particular
fraction, if you carry out
the partial fraction
decomposition, what you get
is right here.
So OK, so those are, that
was a few more examples
of the partial
fraction decomposition.
I hope you enjoyed them,
and I'm going to end there.
