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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
practice integration by parts.

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Specifically, I'd like to
solve the following four

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problems. Or I'd like you to
solve the following four

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problems. I'd like us to find
antiderivatives for each of

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these functions. x e to the
minus x, x cubed over the

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quantity 1 plus x squared
squared, arc tan x, and

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natural log x over x squared.

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And so the main goal, because
we're using integration by

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parts, is to figure out what
you should make u, and what

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you should make v prime.

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And why don't you
give it a shot.

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Work on that for a little bit.

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I'm actually going to give you
one hint, and that's that this

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one, you may want to break up
in a nontraditional way.

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You may not want to break
it up as x cubed

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and 1 over this function.

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You're going to want
to split up this

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function in the numerator.

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Part of it will be in u, part
of it will be in v prime.

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So that's my hint on number 2.

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So now with that information,
I'd like you to give it a

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shot, and then I'll come back,
and I'll show you how I do it.

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OK, welcome back.

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So again, what we're looking
for is antiderivatives for

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each of these four functions.

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And now, what I'm going to do,
is I'm going to help you pick

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u and v prime, and then
I'm going to show you

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what answer I got.

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And I'm going to let you do
the work in the middle.

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So let's start off
with number 1.

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So if I have xe to the minus x
integral of xe to the minus x

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dx, it's very easy to do either,
to make either e to

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the minus x either
u or v prime.

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Doesn't really matter.

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Because an integral of e to the
minus x is going to have

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an e to the minus x again, and
a derivative is going to have

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an e to the minus x again, with
a minus sign in front, in

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both cases.

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But this doesn't
really change.

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So we have, when we go up or
down, it doesn't really matter

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if we integrate up or
take a derivative.

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So really it's, we get to pick
what we do with the c to the

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minus x based on what we
want to do with the x.

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Well, we like taking derivatives
of things that

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don't have two functions of x,
so it would be nice if we

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chose our integration by parts
pieces so that this thing

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wasn't there anymore.

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So let me write down, actually,
before I even do

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number 1, maybe I should
remind you what the

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integration by parts
formula is.

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So let me just, I'll scratch
that out for a second.

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And what we're doing, is we're
going to have integral

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of u v prime dx.

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And if you recall what you saw
in lecture, is this should be

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equal to uv minus the integral
v u prime dx.

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And we'll put that plus c,
because sometimes I forget to

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write it at the end.

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So I'll put it there,
so I don't forget.

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So really, what we're trying
to do, right, is pick the u

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and v prime.

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And so we want to make this
thing, this v u prime, as

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simple as possible.

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So what I was saying is if we
make this v prime or this u,

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it doesn't matter.

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So let's pick whether we want
this to be u or v prime.

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Well, if I make this u,
then u prime is 1.

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That's good.

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If I make it v prime, then
v is x squared over 2.

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That's more complicated.

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So we obviously want
to make this u.

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So for number 1, we're going to
choose u is equal to x, and

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v prime is equal to
e to the minus x.

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And then you can proceed
from there.

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And I'll leave it at that.

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Well, actually, just to make
sure we're OK, I'll even write

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u prime is equal to 1, and
v is going to be equal to

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negative e to the minus x.

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So we'd be able to proceed
from there, right?

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We have all the pieces
we need.

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Now, number 2--

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I'll give you the final
answers at the end.

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Number 2, picking u and v
prime is a little more

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complicated.

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And let's look at
this function.

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x cubed over 1 plus
x squared squared.

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The problem with picking--

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that does not look like a 2.

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Sorry.

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The problem with picking u and
v prime here, is that it's

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hard to see what's going to
be easy to integrate.

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So what we want to do
is rewrite this as--

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let's see-- x squared times x
over 1 plus x squared squared.

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And now, why is this
any better?

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Well, I mean, it's
the same thing.

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But why does this help us
see what we want to do?

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Well, if you notice this
thing right here--

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1 plus x squared.

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What is its derivative?

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Its derivative is 2x.

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Up here we have an x.

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So this piece right here looks
like it could be much more

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easily integrated then
this right here.

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So this might be a little
counterintuitive, because

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we're going to take the
harder-looking thing, and make

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that our v prime.

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But the nice thing is
that we can actually

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integrate this quantity.

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So we choose, in this case,
this is our u, and

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this is our v prime.

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So how do I integrate this?

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Well, I integrate this by
using a substitution.

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And that will give me v. And
the derivative of this is

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quite simple.

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It's just 2x.

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Right?

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But this is the strategy
that we want here.

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Why did we even think to split
that up like that?

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Well, we knew we had to deal
with the denominator in some

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fashion, and taking a derivative
with this in the

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denominator, so putting this
part of the function in u--

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when I look at u prime, it's
going to be even worse.

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It's going to be a higher
power here.

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It's going to be a cubic
in the denominator.

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That's just making
things worse.

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So we know we'd like to
integrate this denominator.

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We'd like it to be a
part of v prime.

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But the problem is that if I put
all the x cubed in the u,

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and if I just had a 1 here for
my v prime, that's, I can't

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really integrate
that very well.

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But if I keep one of the x's,
then I can integrate this

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quite simply with
a substitution.

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So that's the sort of reasoning
behind why we choose

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it that way.

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All right.

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We've got two more to look
at, and then I'll

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give you the answers.

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3.

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OK.

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3, you've seen this
trick before.

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The function was arc tan x.

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Now, you've seen this trick
I'm about to do with

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natural log of x.

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The same kind of thing with
natural log of x.

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You actually saw this in one
of the lecture videos.

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Because there's only one
function here, you might

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think, well, I have no idea what
I'm supposed to pick for

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u and v prime.

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But remember, it's really
arc tan x times 1.

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Now I have two functions.

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And what gives us a hint for why
we would want to do this,

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is that what's the derivative
of arc tan x?

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Let me just remind you.

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d dx of arc tan x is 1 over
1 plus x squared.

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Right?

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We're back to actually an almost
similar situation to

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what we had in the
previous thing.

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d dx of arc tan x is 1 over
1 plus x squared.

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So taking a derivative of this
puts it in a form that almost

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looks easy to integrate.

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What would make this function
easy to integrate?

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If there was an x up here,
instead of a 1.

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Then I could use substitution.

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Where do we get that x from
when we're solving this

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problem, where we're
actually finding an

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antiderivative of arc tan x?

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Well, it's going to come from
the fact that I make this u,

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and I make 1 v prime.

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So let me write that
out explicitly.

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u I make arc tan x, and
v prime I make 1.

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What does that do
in our formula?

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Well, we're going to
be integrating

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something that is v u prime.

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Well, v is going to be x, and
u prime we see right here.

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So it's going to be, I'm going
to be integrating x over 1

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plus x squared when I
started doing the

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integration by parts method.

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That's much simpler, as we
talked about previously,

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because the derivative of x
squared is 2x, and you have an

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x in the numerator when
you put in that v.

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So this is sort of the flavor
of how these things are

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actually working.

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So let me do the
final one here.

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We have ln x over x squared.

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OK.

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Let me just tell
you right now.

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In integration by parts, natural
log x is not something

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you want to make the v prime.

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You don't want to try and
take an antiderivative.

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You know an antiderivative
of natural log of x.

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x ln x minus x.

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But that's certainly not going
to make things any easier.

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Right?

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You're actually, then you've got
a product of two functions

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all of a sudden.

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Everything's getting
more complicated.

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But natural log of x has a very
nice derivative, because

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you end up with something that
has just a power of x.

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Derivative of natural log
of x, just 1 over x.

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So that's probably the way you
always want to go when you see

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natural log of x in these
integration by parts

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techniques.

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Because if I choose u is equal
to ln x, and then v prime.

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In this case, I'm going to
write it as a power.

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Let's think about
what happened.

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u prime is 1 over x, right?

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So u prime is x to
the minus 1.

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What's v?

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Well, it's something
like, let's see.

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Negative x to the minus 1.

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Something like that, right?

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Let's make sure I
did that right.

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Yeah, I think I did
that right.

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So all of a sudden, if I
integrate v u prime, that's

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just a power rule.

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It's x to the minus 2, negative
x to the minus 2.

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So that's quite easy
to integrate.

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So again, when I see natural log
of x in an integration by

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parts method, almost always, I
hate to say always, almost

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always, almost a guarantee
that you want to take a

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derivative.

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You want to make that the u.

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So hopefully that makes sense,
some of these strategies.

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I tried to pick ones that were
somewhat different, so you

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could see some different types
of strategies we needed.

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And now I've done
these earlier.

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So I'm just going to write down
what the answers actually

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are, and you can compare
to what you got.

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So the answer to number
1, just to check.

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Number 2.

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Some of these are
kind of long.

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Number 3.

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Number 4.

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So let's just go through.

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We get, In number 1, we get
negative x e to the minus x

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minus e to the minus x plus c.

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Number 2, we get negative x
squared over 2 times 1 plus x

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squared plus 1/2 natural log
1 plus x squared plus c.

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00:11:01,480 --> 00:11:06,210
3 is x arc tan x minus 1/2
natural log of the quantity 1

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plus x squared plus c, and 4 is
negative natural log x over

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x minus 1 over x plus c.

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So again, the whole point of
this exercise, in my mind, is

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really to make sure we get a
good understanding of, when

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00:11:21,800 --> 00:11:24,440
we're doing integration by
parts, which function makes

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00:11:24,440 --> 00:11:27,115
the most sense to have as u, and
which function makes the

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00:11:27,115 --> 00:11:28,280
most sense to have as v prime.

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So that was the main point
of this exercise.

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Hopefully you're starting to
get a flavor for how these

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00:11:33,770 --> 00:11:35,340
problems actually work.

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00:11:35,340 --> 00:11:37,770
And I think I will stop there.

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