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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 problems.

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

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I'd like us to find
antiderivatives

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for each of these functions.
x e to the minus x,

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x cubed over the quantity
1 plus x squared squared,

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arctan x, and natural
log x over x squared.

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

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

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and what 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,

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and that's that this one,
you may want to break up

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in a nontraditional way.

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You may not want to break
it up as x cubed and 1

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over this function.

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You're going to want to split up
this 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 shot,

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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 each

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of these four functions.

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And now, what I'm
going to do, is

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I'm going to help you
pick u and v prime,

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and then I'm going to show
you 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 x e to the
minus x-- integral of x e

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to the minus x
dx, it's very easy

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to do either-- to make either
e to the minus x either u or v

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prime.

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Doesn't really matter.

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Because an integral
of e to the minus

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x is going to have an
e to the minus x again,

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

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with a minus sign in
front, in 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,

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it doesn't really matter
if we integrate up

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or take a derivative.

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So really it's, we
get to pick what

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we do with the e to the
minus x based on what

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we want to do with the x.

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Well, we like taking
derivatives of things

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

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chose our integration
by parts pieces

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so that this thing
wasn't there anymore.

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

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do number 1, maybe
I should remind you

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what the 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

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integral of u v prime dx.

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And if you recall what
you saw in lecture,

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is this should be equal to u*v
minus the integral v u prime

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dx.

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

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to 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,

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this v u prime, as
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,

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and 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,

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I'll even write u
prime is equal to 1,

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and v is going to be equal
to 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-- 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 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--
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,

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is that it's hard to see what's
going to be easy to integrate.

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So what we want to
do is rewrite this

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as-- let's see-- x
squared times x over 1

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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-- 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

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could be much more easily
integrated than this right

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here.

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So this might be a
little counterintuitive,

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

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and make 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,

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and 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

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is 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

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in some fashion, and
taking a derivative

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with this in the
denominator-- so

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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

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can't really integrate
that very well.

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

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can integrate this quite
simply with a substitution.

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So that's the sort of
reasoning behind why

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we choose 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 arctan x.

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

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with 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,

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you might think, well, I have
no idea what I'm supposed

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to pick for u and v prime.

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But remember, it's
really arctan 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 arctan x?

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Let me just remind you.

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d/dx of arctan 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 what

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we had in the previous thing.

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d/dx of arctan x is 1
over 1 plus x squared.

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So taking a derivative
of this puts it

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in a form that almost
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

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solving this problem, where
we're actually finding

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an antiderivative of arctan 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 arctan 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 integration by parts

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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,

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and you have an x in
the numerator when

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you put in that v.

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

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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

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is not something 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

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a product of two
functions 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 you

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

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

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So that's probably
the way you always

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want to go when you see natural
log of x in these integration

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by parts 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,

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that's 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

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by parts method,
almost always, I

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hate to say always,
almost always,

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almost a guarantee that you
want to take a 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,

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so you 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

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over 2 times 1 plus x squared
plus 1/2 natural log 1 plus x

250
00:10:59,300 --> 00:11:01,480
squared plus c.

251
00:11:01,480 --> 00:11:06,210
Three is x arc tan x minus 1/2
natural log of the quantity 1

252
00:11:06,210 --> 00:11:10,035
plus x squared plus c, and
four is negative natural log

253
00:11:10,035 --> 00:11:15,290
x over x minus 1 over x plus c.

254
00:11:15,290 --> 00:11:18,550
So again, the whole point of
this exercise, in my mind,

255
00:11:18,550 --> 00:11:21,600
is really to make sure we
get a good understanding of,

256
00:11:21,600 --> 00:11:24,440
when we're doing integration
by parts, which function makes

257
00:11:24,440 --> 00:11:27,200
the most sense to have as u, and
which function makes the most

258
00:11:27,200 --> 00:11:28,280
sense to have as v prime.

259
00:11:28,280 --> 00:11:30,470
So that was the main
point of this exercise.

260
00:11:30,470 --> 00:11:33,110
Hopefully you're starting
to get a flavor for how

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00:11:33,110 --> 00:11:35,340
these problems actually work.

262
00:11:35,340 --> 00:11:37,331
And I think I will stop there.

263
00:11:37,331 --> 00:11:37,831