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PROFESSOR: Today we're going to
continue our discussion of
methods of integration.
The method that I'm going to
describe today handles a whole
class of functions of
the following form.
You take P (x) / Q (x)
and this is known
as a rational function.
And all that means is that
it's a ratio off two
polynomials, which are these
functions P ( x) and Q ( x).
We'll handle all such functions
by a method which is
known as partial fractions.
And what this does is, it splits
P / Q into what you
could call easier pieces.
So that's going to be some
kind of algebra.
And that's what we're going
to spend most of
our time doing today.
I'll start with an example.
And all of my examples
will be illustrating
more general methods.
The example is to integrate the
function 1 / x - 1 +, say,
3 / x + 2 dx.
That's easy to do.
It's just, we already
know the answer.
It's ln x - 1 + 3 ln x + 3.
Plus a constant.
So that's done.
So now, here's the difficulty
that is going to arise.
The difficulty is that I can
start with this function,
which is perfectly manageable.
And than I can add these
two functions together.
The way I add fractions.
So that's getting a common
denominator.
And so that gives me x +
2 here + 3 ( x - 1).
And now if I combine together
all of these terms, then
altogether I have 4x +
2 - 3, that's - 1.
And if I multiply out the
denominator that's x ^2 + that
2 turned into a 3, that's
interesting.
Hope there aren't too many more
of those transformations.
Is there another one here?
STUDENT: [INAUDIBLE]
PROFESSOR: Oh, it happened
earlier on.
Wow that's an interesting
vibration there.
OK.
Thank you.
So, I guess my 3's were
speaking to my 2's.
Somewhere in my past.
OK, anyway, I
think this is now correct.
So the problem is
the following.
This is the problem with this.
This integral was easy.
I'm calling it easy, we already
know how to do it.
Over here.
But now over here,
it's disguised.
It's the same function, but it's
no longer clear how to
integrate it.
If you're faced with this
one, you say what am
I supposed to do.
And we have to get around
that difficulty.
And so what we're going
to do is we're going
to unwind this disguise.
So we have the algebra problem
that we have. Oh, wow.
There must be something
in the water.
Impressive.
Wow.
OK, let's see.
Is 2/3 = 3/2?
Holy cow.
Well that's good.
Well, I'll keep you awake
today with several other
transpositions here.
So our algebra problem is to
detect the easy pieces which
are inside.
And the method that we're going
to use, the one that
we'll emphasize anyway, is one
algebraic trick which is a
shortcut, which is called
the cover-up method.
But we're going to talk
about even more
general things than that.
But anyway, this is where
we're headed.
Is something called the
cover-up method.
Alright.
So that's our intro.
And I'll just have to remember
that 2 is not 3.
I'll keep on repeating that.
So now here I'm going to
describe to you how we unwind
this disguise.
The first step is,
we write down the
function we want to integrate.
Which was this.
And now we have to undo the
first damage that we did.
So the first step is to factor
the denominator.
And that factors, we happen to
know the factors, so I'm not
going to carry this out.
But this can be a rather
difficult step.
But we're going to assume
that it's done.
For the purposes of
illustration here.
So I factor the denominator.
And now, the second thing that
I'm going to do is what I'm
going to call the setup here.
How I'm going to
set things up.
And I'll tell you what these
things are more systematically
in a second.
And the setup is that
I want to somehow
detect what I did before.
And I'm going to write
some unknowns here.
What I expect is that this will
break up into two pieces.
One with the denominator x -
1, and the other with the
denominator x + 2.
So now, my third step is going
to be to solve for A and B.
And then I'm done,
if I do that.
That's the complete unwinding
of this disguise.
And this is where the cover-up
method comes in handy.
This is this method that
I'm about to describe.
Now, you can do the algebra in a
clumsy way, or you can do it
in a quick way.
And we'd like to get efficient
about the algebra involved.
And so let me show you
what the first step
in the trick is.
We're going to solve for A by
multiplying by (x - 1).
Now, notice if you multiply by
(x - 1) in that equation 2,
what you get is this.
You got 4x - 2 / the
x - 1's cancel.
You get this on the
left-hand side.
And on the right-hand side
you get A. The x
- 1's cancel again.
And then we get this
extra term.
Which is B/ ( x + 2)( x - 1).
Now, the trick here, and we're
going to get even better trick
in just a second.
The trick here is that I
didn't try to clear the
denominators completely.
I was very efficient about
the way I did it.
It just cleared one factor.
And the result here
is very useful.
Namely, if I plug in now x =
1, this term drops out too.
So what I'm going to do now is
I'm going to plug in x = 1.
And what I get on the left-hand
side here is 4 - 1
and 1 + 2, and on the
left-hand side I get
A. That's the end.
This is my formula for A. A
happens to be equal to 1.
And that's, of course,
what I expect.
A had better be 1, because the
thing broke up into 1 / x - 1
+ 3 / x + 2.
So this is the correct answer.
There was a question out
here, which I missed.
STUDENT: Aren't polynomials
defined as functions with
whole powers, or could
they be square roots?
PROFESSOR: Are polynomials
defined as functions with
whole powers, or can they
be square roots?
That's the question.
The answer is, they only
have whole powers.
So for instance here I only
have the power 1 and 0.
Here I have the powers 2, 1
and 0 in the denominator.
Square roots are no good
for this method.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Why did
I say x = 1?
The reason why I said x = 1 was
that it works really fast.
You can't know this in advance,
that's part of the method.
It just turns out to be
the best thing to do.
The fastest way of getting at
the coefficient A. Now the
curious thing, let me
just pause for a
second before I do it.
If I had plugged x = 1 into the
original equation, I would
have gotten nonsense.
Because I would've gotten
0 in the denominator.
And that seems like the most
horrible thing to do.
The worst possible thing
to do, is to set x = 1.
On the other hand, what
we did is a trick.
We multiplied by x - 1.
And that turned the equation
into this.
So now, in disguise,
I multiplied by 0.
But that turns out
to be legitimate.
Because really this equation
is true for all x except 1.
And then instead of taking
x = 1, I can really
take x tends to 1.
That's really what I need.
The limit is x goes to one.
The equation is still
valid then.
So I'm using the worst case, the
case that looks like it's
dividing by 0.
And it's helping me because it's
cancelling out all the
information in terms of B. So
the advantage here is this
cancellation that occurs
in this part.
So that's the method.
We're going to shorten it much,
much more in a second.
But let me carry it out for the
other coefficient as well.
So the other coefficient I'm
going to solve for B, I'm
going to multiply by x + 2.
And when I do that, I get 4x
- 1 / x - 1, that's the
left-hand side, the very
top expression there.
And then down below I get
A/ ( x - 1)( x + 2).
And then again the
x + 2's cancel.
So I get B sitting alone.
And now I'm going to
do the same trick.
I'm going to set x = - 2.
That's the value which
is going to knock
out this A term here.
So that cancels this
term completely.
And what we get here all told
is - 8 - 1 / - 2 - 1 = B. In
other words, B = 3, which
was also what it
was supposed to be.
B was this number
3, right here.
Which I'm now not going
to change to 2.
Because I know that
it's not 2.
There was a question.
STUDENT: [INAUDIBLE]
PROFESSOR: All right.
Now, this is the method which
is called cover-up.
But it's really carried out
much, much faster than this.
So I'm going to review the
method and I'm going to show
you what it is in general.
So the first step is to factor
the denominator, Q. That's
what I labeled 1 over there.
That was the factorization of
the denominator up top.
The second step is what I'm
going to call the setup.
That's step 2.
And that's where I knew what I
was aiming for in advance.
And I'm going to have to
explain to you in every
instance exactly what this
setup should be.
That is, what the unknowns
should be and what target,
simplified expression,
we're aiming for.
So that's the setup.
And then the third step is what
I'll now call cover-up.
Which is just a very fast way
of doing what I did on this
last board, which is solving for
the unknown coefficients.
So now, let me perform
it for you again.
Over here.
So it's 4x - 1 divided
by, so this is to
eliminate writing here.
Handwriting it makes
it much faster.
So this part just factoring the
denominator, that was 1,
that was step 1.
And then step 2, again,
is the setup, which is
setting it up like this.
Alright, that's the setup.
And now I claim that without
writing very much, I can
figure out what A and B are.
Just by staring at this.
So now what I'm going to do is
I'm just going to think what I
did over there.
And I'm just going to
do it directly.
So let me show you what the
method consists of visually.
I'm going to cover up, that is,
knock out this factor, and
focus on this number here.
And I'm going to plug in the
thing that makes the 0,
which is x = 1.
So I'm plugging in x = 1.
To this left-hand side.
And what I get is 4 - 1 / 1 +
2 = A. Now, that's the same
thing I did over there.
I just did it by skipping the
intermediate algebra step,
which is a lot of writing.
So the cover-up method
really amounts to
the following thing.
I'm thinking of multiplying
this over here.
It cancels this and it gets
rid of everything else.
And it just leaves me with
A on the right-hand side.
And I have to get rid
of it on this side.
So in other words, by
eliminating this, I'm
isolating a on the
right-hand side.
So the cover-up is that
I'm covering this and
getting A out of it.
Similarly, I can do the same
thing with B. It's focused on
the value x = - 2.
And b is what I'm getting
on the right-hand side.
And then I have to
cover up this.
So if I cover up that, then
what's left over with x = - 2
is again, - 8 - 1 / - 2 - 1.
So this is the way the method
gets carried out in practice.
Writing, essentially,
the least you can.
Now, when you get to several
variables, it becomes just way
more convenient to do this.
So now, let me just review
when cover-up works.
So this cover-up method
works if Q ( x) has
distinct linear factors.
And, so you need two
things here.
It has to factor completely, the
denominator has to factor
completely.
And the degree of the numerator
has to be strictly
less than the degree
of the denominator.
I'm going to give you
an example here.
So, for instance, and this tells
you the general pattern
of the setup also.
Say you had x ^2 + 3x
+ 8, let's say.
Over (x - 1) ( x - 2)( x + 5).
So here I'm going to
tell you the setup.
The setup is going to be
A / (x - 1) + B / (x -
2) + c / x + 5.
And it will always break
up into something.
So however many factors you
have, you'll have to put in a
term for each of those.
And then you can find
each number here by
this cover-up method.
Now we're done with that.
And now we have to go on to the
algebraic complications.
So would the first typical
algebraic complication be.
It would be repeated roots
or repeated factors.
Let me get one that doesn't
come out to be
extremely ugly here.
So this is what we'll
call Example 2.
And this is going to work when
the degree, you always need
that the degree of the numerator
is less than the
degree of the denominator.
And Q has now repeated
linear factors.
So let's see which example
I wanted to show you.
So let's just give this here.
I'll just repeat the
denominator.
With an extra factor
like this.
Now, the main thing you need
to know, since I've already
performed the factorization
for you.
Already performed Step 1.
This is Step 1 here.
You have to factor things all
the way, and that's already
been done for you.
And here's what this setup is.
The setup is that it's of
the form A / (x - 1)
+ B / (x - 1) ^2.
We need another term for the
square here. + C / (x + 2).
In general, if you have more
powers you just need to keep
on putting in those powers.
You need one for each
of the powers.
Why does it have to be squared?
OK.
Good question.
So why in the world
am I doing this?
Let me just give you one hint
as to why I'm doing this.
It's very, very much like the
decimal expansion of a number
or, say, the base 2 expansion
of a number.
So, for, example the number 7/16
is 0 / 2 + 1 / 2 ^2 + 1/2
^3 +, is that right?
So it's 4/16 + 1/2 ^4.
It's this sort of thing.
And I'm getting this power
and this power.
If I have higher powers,
I'm going to have to
have more and more.
So this is what happens
when I have a 2 ^ 4.
I have to represent
things like this.
That's what's coming out
of this piece with the
repetitious here.
Of the powers.
This is just an analogy.
Of what we're doing.
Yeah, another question
over here.
STUDENT: [INAUDIBLE]
PROFESSOR: Yes.
So this is an example, but it's
meant to represent the
general case and I will also
give you a general picture.
For sure, once you have the
second power here, you'll need
both the first and the second
power mentioned over here.
And since there's only a first
power over here I only have to
mention a first power
over there.
If this were a 3 here, there
would be one more term which
would be the one for x - 1
^2 in the denominator.
That's what you just said.
OK, now, what's different about
this setup is that the
cover-up method, although
it works, it
doesn't work so well.
It doesn't work quite as well.
The cover-up works for the
coefficients B and C, not A.
We'll have a quick method for
the numbers B and C. To figure
out what they are.
But it will be a little slower
to get to A, which we will do
last. Let me show you
how it works.
First of all, I'm going to do
the ordinary cover-up with C.
So for C, I just want
to do the same old
thing that I did before.
I cover up this, and that's
going to get rid of all the
junk except for the C term.
So I have to plug x = - 2.
And I get x -- sorry, I get (-2
) ^2 + 2 in the numerator.
And I get (- 2 - 1)^2
in the denominator.
Remember I'm covering this up.
So that's all there is on
the left-hand side.
And on the right-hand side all
there is C. Everything else
got killed off, because it
was x - 2 times that.
That's 0 times all
that other stuff.
And the x - 2 over
here canceled.
This is the shortcut that I just
described, and this is
much faster than doing
all that arithmetic.
And algebra.
So all together this
is a 6/9, right?
So it's C = 6/9, which is 2/3.
Now, the other one which is easy
to do, I'm going to do by
the slow method first.
But you omit a term.
The idea is to cover up
the other bad factor.
Cover ups, I'll do it both
the way and the slow way.
I'll do it the fast way first,
and then I'll show
you the slow way.
The fast way is to
cover this up.
And then I have to cover
up everything else.
That gets eliminated.
And that includes everything
but B. So I
get B on this side.
And I get 1 on that side.
So that's 1 ^2 + 2 / 1 + 2.
So in other words, B = 1.
That was pretty fast, so let
me show you what arithmetic
was hiding behind that.
What algebra was hiding
behind it.
What I was really
doing is this.
In multiplying through by
x - 1 ^2, so I got this.
So this canceled here, so this
C just stands alone.
And then I have here C
/x + 2 (x - 1) ^2.
Notice again, I cleared out
that 1, this term from the
denominator and sent it over
to the other side.
Now, what's happening is that
when I set x = 1 here, this
term is dying.
This term is going away, because
there's more powers in
the numerator than in
the denominator.
This is still 0.
And this one is gone also.
So all that's left is B. Now, I
cannot pull that off with a
single power of x - 1.
I can't expose the A term.
It's the B term that
I can expose.
Because I can multiply through
by this thing squared.
If I multiply through by just x
- 1, what'll happen here is
I won't have canceled
this (x - 1 )^2.
It's useless.
I still have a 0 in
the denominator.
I'll have B / 0 when
I plug in x = 1.
Which I can't use.
Again, the cover-up method is
giving us B and C, not A.
Now, for the last term, for A,
I'm going to just have to be
straightforward about it.
And so I'll just suggest
for A, plug in
your favorite number.
So plug in my favorite number.
Which is x = 0.
And you won't be able to
plug in x = 0 if you've
already used it.
Here the two numbers we've
already used are x
= 1 and x = - 2.
But we haven't used x =
0 yet, so that's good.
I'm going to plug in now x
= 0 into the equation.
What do I get?
I get 0 ^2 + 2 / (- 1) ^2 *
2 is equal to, let's see.
A is the thing that
I don't know.
So it's A / - 1 +, B /
x - 1 ^2 so B = 1, so
that's 1 / (- 1) ^2.
And then C was 2/3.
2/3 / x + 2.
So that's 0 + 2.
Don't give up at this point.
This is a lot of algebra.
You really have to plug
in all these numbers.
You make one arithmetic mistake
and you're always
going to get the wrong answer.
This is very arithmetically
intensive.
However, it does simplify
at this point.
We have 2 / 2, that's 1.
Is equal to - A + 1 + 1/3.
So let's see.
A on the other side, this
becomes A = 1/3.
And that's it.
This is the end.
We've we've simplified
our function.
And now it's easy to integrate.
Question.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: So the question is,
if x = 0 has already been
used, what do I do?
And the answer is, pick
something else.
And you said pick
a random number.
And that's right, except that if
you really picked a random
number it would be 4.12567843,
which would be difficult.
What you want to pick is the
easiest possible number you
can think of.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: If you had, as in
this sort of situation here.
More powers.
Wouldn't you have
more variables.
Very good question.
That's absolutely right.
This was a 3 by 3 system in
disguise, for these three
unknowns, A, B and C. What we
started with in the previous
problem was two variables.
It's over here, the variables A
and B. And as the degree of
the denominator goes up, the
number of variables goes up.
It gets more and more and
more complicated.
More and more arithmetically
intensive.
STUDENT: [INAUDIBLE]
PROFESSOR: Well, so.
The question is, how would
you solve it if
you have two unknowns.
That's exactly the point here.
This is a system
of simultaneous
equations for unknowns.
And we have little tricks for
isolating single variables.
Otherwise we're stuck with
solving the whole system.
And you'd have to solve the
whole system by elimination,
various other tricks.
I'll say a little more
about that later.
Now, I have to get one
step more complicated
with my next example.
My next example is going to
have a quadratic factor.
So still I'm sticking to the
degree of the polynomial and
the numerator is less than the
degree of the polynomial in
the denominator.
And I'm going to take
the case where Q
has a quadratic factor.
Let me just again illustrate
this by example.
I have here (x - 1) (x^2 + 1).
I'll make it about as
easy as they come.
Now, the setup will be slightly
different here.
Here's the setup.
It's already factored.
I've already done as
much as I can do.
I can't factor this x^2 + 1 into
linear factors unless you
know about complex numbers.
If you know about complex
numbers this method becomes
much easier.
And it comes back to the
cover-up method.
Which is the way that the
cover-up method was originally
conceived by heavy side.
But you won't get to
that until 18.03.
So we'll wait.
This, by the way, is a method
which is used for integration.
But it was invented to do
something with Laplace
transforms and inversion
of certain kinds of
differential equations.
By heavy side.
And so it came much later
than integration.
But anyway, it's a very
convenient method.
So here's the set up
with this one.
Again, we want a term for
this (x - 1) factor.
And now we're going to also
have a term with the
denominator x squared plus 1.
But this is the difference.
It's now going to be a first
degree polynomial.
One degree down from
the quadratic here.
So this is what I keep
on calling the setup,
this is number 2.
You have to know that in advance
based on the pattern
that you see on the
left-hand side.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is, if
the degree of the numerator.
So in this case, if this were
cubed, and this is matching
with the denominator, which
is total of degree 3.
The answer is that this
does not work.
STUDENT: [INAUDIBLE]
PROFESSOR: It definitely
doesn't work.
And we're going to have to
do something totally
different to handle it.
Which turns out, fortunately,
to be much easier than this.
But we'll deal with
that at the end.
Keep this in mind.
This is an easy way to make a
mistake if you start with a
higher degree numerator.
You'll never get the
right answer.
So now, so I have
my setup now.
And now what can I do?
Well, I claim that I can still
do cover-up for A.
It's the same idea.
I cover this guy up.
And if I really multiply by it
it would knock everything out
but A. So I cover this up
and I plug in x = 1.
So I get here 1 ^2 / 1 ^2 + 1 =
A. In other words, A = 1/2.
Again cover-up is pretty
fast, as you can see.
It's not too bad.
Now, at this next stage, I want
to find B and C. And the
best idea is the slow way.
Here, it's not too terrible.
But it's just what we're
going to do.
Which is to clear the
denominators completely.
So for B and C, just clear
the denominator.
That means multiply through
by that whole business.
Now, when you do that on the
left-hand side you're
going to get x ^2.
Because I got rid of the
whole denominator.
On the right-hand side when I
bring this up, the x - 1 will
cancel with this.
So the a term will
be A ( x ^2 + 1).
And the Bx + C term will have
a remaining factor of x - 1.
Because the x ^2 +
1 will cancel.
Again, the arithmetic here
is not too terrible.
Now I'm going to do
the following.
I'm going to look at
the x ^2 term.
On the left-hand side and
the right-hand side.
And that will give me one
equation for B and C. And then
I'm going to do the same thing
with another term.
The x^2 term on the left-hand
side, the coefficient is 1.
It's 1 ( x ^2).
On the other side, it's A.
remember I actually have A. So
I'm going to put it
in, it's 1/2.
So this is the A term.
And so I get 1/2 ( x ^2).
And then the only other
x^2 is when this Bx
multiplies this x.
So Bx * x is Bx ^2, so this is
the other coefficient on x ^2
is B. And that forces
B to be 1/2.
And last of all, I'm going to
do the x^ 0 power term.
Or, otherwise known as
the constant term.
And on the left-hand side,
the constant term is 0.
There is no constant term.
On the right-hand side there's
a constant term, 1/2 * 1.
That's 1/2 here.
And then there's another
constant term, which is this
constant times this - 1.
Which is - C. And so
the conclusion
here is that C = 1/2.
Another question.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: There's also an x^
0 power hidden in here.
Sorry, an x ^ 1 , that's what
you were asking about, sorry.
There's also an x ^ 1 .
The only reason why I didn't
go to the x^ 1 is that it
turns out with these two
I didn't need it.
The other thing is that by
experience, I know that the
extreme ends of the
multiplication
are the easiest ends.
And the middle terms have tons
of cross terms. And so I don't
like the middle term as much
because it always involves
more arithmetic.
So I stick to the lowest and
the highest terms if I can.
So that was really
a sneaky thing.
I did that without
saying anything.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Another
good question.
Could I just set equals 0?
Absolutely.
In fact, that's equivalent to
picking out the x^ 0 term.
And you could plug in numbers.
If you wanted.
That's another way of doing
this besides doing that.
So you can also plug
in numbers.
Can plug in numbers. x = 0.
Actually, not x = 1,
right? - 1, 2, etc.
Not 1 just because we've
already used it.
We won't get interesting
information out.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: So the question
is, could I have done
it this other way.
With the polynomial, with
this other one.
Yes, absolutely.
So in other words what I've
taught you now is two choices
which are equally reasonable.
The one that I picked was the
one that was the fastest for
this guy and the one that was
fastest for this one, but I
could've done the other
way around.
There are a lot of
ways of solving
simultaneous equations.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is the following.
So now everybody can understand
the question.
If this, instead of being x
^2 + 1, this were x^3 + 1.
So that's an important
case to understand.
That's a case in which
this denominator
is not fully factored.
So it's an x^3 + 1, you would
have to factor out an x + 1.
So that would be a situation
like this, you have an x^3 +
1, but that's (x + 1)( x^2 +
x + 1), this kind of thing.
If that's the right, there
must be a minus
sign in here maybe.
OK, something like this.
Right?
Isn't that what it is?
STUDENT: [INAUDIBLE]
PROFESSOR: I think it's right.
But anyway, the point is that
you have to factor it.
And then you have a linear
and a quadratic.
So you're always going to be
faced eventually with linear
factors and quadratic factors.
If you have a cubic, that means
you haven't factored
sufficiently.
So you're still back
in Step 1.
STUDENT: [INAUDIBLE]
PROFESSOR: In the
x^3 + 1 case?
STUDENT: [INAUDIBLE]
PROFESSOR: In the x^3 + 1 case,
we are out of luck until
we've completed the
factorization.
Once we've completed the
factorization, we're going to
have to deal with these two
factors as denominators.
So it'll be this plus something
over x + 1 + a Bx +
C type of thing over
this thing here.
That's what's eventually
going to happen.
But hold on to that idea.
Let me carry out one
more example here.
So I've figured out what
all the values are.
But I think it's also worth it
to remember now that we also
have to carry out
the integration.
What I've just shown you is that
the integral of x ^2 dx
over (x - 1)( x ^2 + 1) is equal
to, and I've split up
into these pieces.
So what are the pieces?
The pieces are, 1/2, x -
1 + 1/2 x / x ^2 + 1.
This is the A term.
This is the B term.
And then there's the C term.
So we'd better remember
that we know how to
antidifferentiate
these things.
In other words, I want to
finish the problem.
The others were pretty easy, so
I didn't bother to finish
my sentence, but here I want
to be careful and have you
realize that there's something
a little more to do.
First of all we have the, the
first one is no problem.
That's this.
The second one actually
is not too bad either.
This is, by the advanced
guessing method, my favorite
method, something like the
logarithm, because that's
what's going to appear
in the denominator.
And then, if you differentiate
this, you're going
to get 2x over this.
But here we have 1/2.
So altogether it's
1/4 of this.
So I fixed the coefficient
here.
And then the last one, you have
to think back to some
level of memorization here and
remember that this is 1/2 the
arc tangent.
STUDENT: [INAUDIBLE]
PROFESSOR: Why did
I go to 1/4?
Because in disguise, for this
guy, I was thinking d / dx of
ln (x ^2 + 1) = 2x / x^2 + 1.
Because it's the derivative
of this divided by itself.
This is the derivative
of ln u is u ' / u.
Ln u' = u' / u.
That was what I applied.
And what I had was 1/2, so I
need a total of 1/4 to cancel.
So 2/4 is 1/2.
Now I've got to get you out
of one more deep hole.
And I'm going to save the
general pattern for next time.
But I do want to clarify
one thing.
So let's handle this thing.
What if the degree of P is
bigger than or equal to the
degree of Q. That's the thing
that I claimed was easier.
And I'm going to describe
to you how it's done.
Now, in analogy, with numbers
you might call this an
improper fraction.
That's the thing that should
echo in your mind when you're
thinking about this.
And I'm just going to do
it by example here.
Let's see., I cooked up an
example so that I don't make
an arithmetistic mistake
along the way.
So there are two or three steps
that I need to explain.
So here's an example.
The denominator's degree 2,
the numerator is degree 3.
It well exceeds, so there's
a problem here.
Our method is not
going to work.
And the first step that I
want to carry out is to
reverse Step 1.
That is, I don't want the
factorization for what I'm
going to do next.
I want it multiplied out.
That means I have to multiply
through, so I
get x ^2 + x - 2.
I'm back to the starting
place here.
And now, the next thing that I'm
going to do is, I'm going
to use long division.
That's how you convert
an improper
fraction to a proper fraction.
This is something you were
supposed to learn in, I don't
know, Grade 4, I know.
Grade 3, Grade 4, Grade
5, Grade 6, etc.
So here's how it works in
the case of polynomials.
It's kind of amusing.
So we're dividing this
polynomial into that one.
And so the quotient to first
order here is x.
That is, that's going to match
the top terms. So I get
x^3 + x ^2 - 2x.
That's the product.
And now I subtract.
And it cancels.
So we get here - x ^2 + 2x.
That's the difference.
And now I need to divide
this next term in.
And I need a - 1.
So - 1 times this is
- x ^2 - x + 2.
And I subtract.
And the x^2's cancel.
And here I get + 3x - 2.
Now, this thing has a name.
This is called the quotient.
And this thing also
has a name.
This is called the remainder.
And now I'll show you how it
works by sticking it into the
answer here.
The quotient is x - 1.
And the remainder is, let's
get down there.
3x - 2 / x ^2 + x - 2.
So the punchline here
is that this
thing is easy to integrate.
This is easy.
And this one, you can use, now
you can use cover-up, The
method that we had before.
Because the degree of the
numerator is now below the
degree of the denominator.
It's now first degree and
this is second degree.
What you can't do is
use cover-up to
start out with here.
That will give you
the wrong answer.
So that's the end for today,
and see you next time.
