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Hi.

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Welcome back to recitation.

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In lecture, you've learned a
whole bunch of different

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integration techniques.

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So I have here a couple of
problems that will give you an

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opportunity to apply those
techniques, and figure out

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which ones to apply.

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So in particular I have one
definite integral, the

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integral 0 to 2 of the quantity
x times e to the 1

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minus x squared with
respect to x.

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And the second one is an
indefinite integral of 2 arc

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tan x divided by x squared
with respect to x.

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So why don't you pause the
video, take some time to work

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these out, come back, and we
can work them out together.

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Welcome back.

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Hopefully you had some luck
working on these integrals.

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Let's get started.

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We can do the first one first,
and then we'll go on to the

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second one.

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So here we want to compute the
integral from 0 to 2 of x e to

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the 1 minus x squared dx.

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So in order to do this,
we have to, you know--

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well, we could look at it and
say, do I know the answer

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immediately, off the
top of my head?

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And this is not one of the, you
know, ones that we have

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immediately already memorized
a formula for.

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So then you say, OK.

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So now I need to do something
in order to try

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to integrate it.

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And so, you know, you have a
whole bunch of techniques.

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And for this one, when you look
at it, the technique that

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is going to work for you is just
the simplest technique

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that you have, which is just a
straightforward substitution.

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And one way to figure that out,
is that you can look at

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this and you can see that here
we have a composition of

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functions, this e to the 1
minus x squared, and so

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whenever you have a composition
of functions,

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there's the opportunity for
there to have been a chain

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rule somewhere.

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And in this case, we see that
this composition of functions

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is multiplied by x, and x is
very closely related to the

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derivative of 1 minus
x squared.

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So that's what's suggesting for
us a nice substitution.

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And so the substitution we're
going to try, then, is that

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we're going to put this inside
function, this 1 minus x

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squared, we're going to
set u equal to that.

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So, all right.

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So this is equal to-- so
I'm going to say, u

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equal 1 minus x squared.

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So now if I take a differential,
if u equals 1

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minus x squared, that means
that du is minus 2x dx.

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And since this is a definite
integral, I also need to

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change the bounds
of integration.

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So when x is equal to 0, that
means that u is equal to,

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well, 1 minus 0 squared is 1,
so that's the lower bound.

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And when x is equal to 2, that
means that u is equal to 1

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minus 2 squared, which
is negative 3.

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So we make these
substitutions.

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OK.

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So what does our integral
become?

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So this becomes the integral--

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so now the lower bound becomes
u equals 1, and the upper

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bound is u equals negative 3.

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OK.

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So now e to the 1 minus
x squared, that's

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just e to the u.

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And x dx--

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well, that's almost du.

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It's du divided by negative 2.

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So du divided by negative 2.

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So we make the substitution.

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We transform our integral
to this form.

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Now, this is a very nice,
simple integral, right?

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This is just the integral of e
to the u with a, you know, a

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constant multiple.

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So, OK.

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So we can do this right away.

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So this is--

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I'll pull the minus
1/2 out in front.

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And now fundamental theorem
of calculus.

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Antiderivative of the e to
the u is just e to the u.

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And I have to take that
between my bounds.

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So between u equals 1 and
u equals minus 3.

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So notice that because we
changed bounds, we don't have

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to go back and convert
everything back to x.

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We can just plug in directly.

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So OK.

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So we plug in directly.

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What do I get?

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So first I get minus
1/2 times--

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all right, so I'll just pull
the constant out in front.

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So the first term is e to the
minus third minus e to the 1.

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OK.

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And if I wanted to, I could
rewrite this as e minus e to

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the minus 3 over 2, or in
a variety of other ways.

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All right.

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So this one was a pretty
straightforward integration by

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substitution, definite integral,
change bounds, and

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at the end you get
this answer.

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So let's take a look
at the second one.

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So the second one is a bit
trickier, I think, and it

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requires a bit more work.

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So here it is.

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Let me rewrite it.

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So it's the antiderivative of 2
arc tan x over x squared dx.

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So we have here an
antiderivative

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that we have to compute.

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And this is a kind of
messy-looking expression that

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we're working with.

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All right?

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So this is not very pleasant.

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So there are a couple of
different things we could

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think to do.

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So one thing that you could
think to do is you could say,

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I really don't like arc tan.

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I'm going to try and
get rid of arc tan.

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And the way I'm going to get rid
of arc tan is by doing a

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substitution, u equals arc
tan x, or x equals tan u.

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So if you do that, that'll
get rid of your inverse

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trigonometric function, but
it'll introduce, in the

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bottom, you have this x squared,
so that'll introduce

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like a tan u squared,
and dx will give you

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some more trig functions.

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So what you'll be left
with, then-- so this

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will be u on top.

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So you'll have u times a product
of trig functions.

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So I guess it's up to you
whether you think that a

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polynomial, or a power of
u times a bunch of trig

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functions, is simpler
than an inverse trig

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function times a power.

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It's not clear to me that it's
actually that much simpler.

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And in any case, we wouldn't
sort of be

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finished at that stage.

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There would still be a fair
amount more work to do.

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So we can think about, what
other things could we do?

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That's kind of the only obvious
substitution here.

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But another thing that suggests
itself, is that we

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could try an integration
by parts.

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And one reason is, this is
a product of things.

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You know, here it's arc 10x
times 1 over x squared.

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And arc 10x is something that
works nicely with integration

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by parts if you can
differentiate it.

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Right?

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So inverse trigonometric
functions behave like

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logarithms with respect to
integration by parts.

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By which I mean, they really
like to be differentiated.

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Because when you differentiate
an arc tan, you get something

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much simpler.

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Well, simpler, anyhow.

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Maybe not much simpler.

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You get 1 over 1
plus x squared.

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So if you can apply integration
by parts and

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differentiate the arc tan, that
makes your life nice.

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And so, then, in order for that
to work, you need for

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everything else to be
something you can

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antidifferentiate, and here
everything else is just 2 over

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x squared, and so that's
something that

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we know how to integrate.

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It's easy to do.

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So, OK.

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So let's give that
a try, then.

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So we're going to do integration
by parts, so I'm

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going to set you equal
to arc tan x.

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So that means u prime is equal
to 1 over 1 plus x squared.

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And I'm going to let v prime
equal, well, I guess I have to

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pick up the 2 somewhere, so I
might as well pick it up here.

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2 over x squared.

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And so then v, so I integrate
2 over x squared, so that

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gives me x to the minus
1, with a minus sign.

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So it's minus 2 over x.

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So these are u, u prime,
v and v prime.

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So now OK, so now I apply
integration by parts, and that

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tells me this is equal to--

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well, let's see.

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So it's uv, which is minus
2 arc tan x over x.

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That's uv. Minus the integral
of, now what the second part

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is, v u prime dx.

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So here v u prime is--

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well, OK.

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So they're minus 2.

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I'm going to pull
that out front.

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So it's plus 2--

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all right.

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And then the rest of it
is 1 over x times 1

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plus x squared dx.

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So if you apply integration by
parts, you have the uv stuff

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that gets kicked out out front,
and then what you're

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left with is this
second integral.

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Now this is, to me this is
simpler-looking than what we

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started with.

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It's still not simple,
but it's simpler.

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So it seems to me that, you
know, having reached this

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stage, we can say, OK.

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We've made some progress, and
now we have to keep going.

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Right?

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So it's not obvious to me what
the antiderivative of this

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expression should be.

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So how can I figure that out?

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Well, there are a couple of
different options here.

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One thing you might look
at this and see,

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is you might see--

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again, I have a 1 plus x squared
in the denominator,

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and so that might make you
tempted to do a trig

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substitution again.

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And in fact, you could do a
trig substitution here.

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You could put x equal tan theta,
and you would be able

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to solve the question
like this.

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But I don't think it's the
simplest way to go.

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Because in addition--

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well, on the one hand, this has
this 1 plus x squared in

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the denominator, but on the
other hand, this is just a

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rational function.

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Yeah?

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It's a ratio of two
polynomials.

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The polynomial on top is just
1, and the polynomial on the

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bottom is this product, x
times 1 plus x squared.

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And so whenever you have a
rational function that you're

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trying to integrate, you
can also always use--

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what's it called--

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partial fraction
decomposition.

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So we can try to use partial
fractions on the second

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expression.

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And so I think I'd like
to do it that way.

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So let's do that.

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Let's do it down here.

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So partial fractions on
the quantity 1 over x

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times 1 plus x squared.

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So if you remember your partial
fractions here, this

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is completely factored.

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This quadratic doesn't factor.

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It doesn't have any real
roots, doesn't factor.

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So OK, so we have a single
linear term, and a single

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non-factorable quadratic term.

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So partial fractions tells us
that we can write this in the

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form a over x plus-- and now
because the second term is

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quadratic, it needs to be bx
plus c over 1 plus x squared.

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Right?

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So when you have a quadratic
in the bottom, you get two

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constants up at the top,
a linear polynomial.

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And then you can always check.

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You have three constants here,
and the degree of the

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denominator is 3, so that's
one way of checking that

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you're not completely off base,
that the degree down

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here should always agree with
the number of constants that

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you're solving for.

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OK.

250
00:10:53,530 --> 00:10:53,880
Good.

251
00:10:53,880 --> 00:10:57,310
So now we need to solve
this for a, b, and c.

252
00:10:57,310 --> 00:11:00,310
And so OK, because we have
this nice single linear

253
00:11:00,310 --> 00:11:02,730
factor, we can do the cover
up method there.

254
00:11:02,730 --> 00:11:08,360
So we cover up x, and we cover
up over x, and we cover up

255
00:11:08,360 --> 00:11:09,430
everything else.

256
00:11:09,430 --> 00:11:12,660
So on the right hand side, we
just end up with a, and on the

257
00:11:12,660 --> 00:11:14,290
left hand side, well, so x.

258
00:11:14,290 --> 00:11:16,500
So that's what we need, we
get whatever when we

259
00:11:16,500 --> 00:11:17,960
plug in x equals 0.

260
00:11:17,960 --> 00:11:21,900
So on this side, we get 1
over 1 plus 0 squared,

261
00:11:21,900 --> 00:11:25,060
so that's just 1.

262
00:11:25,060 --> 00:11:29,770
So the cover up method gives
us that a is equal to 1.

263
00:11:29,770 --> 00:11:31,630
And now we need to solve
for b and c.

264
00:11:31,630 --> 00:11:33,910
And probably the most
straightforward way, in this

265
00:11:33,910 --> 00:11:36,830
case, to solve for b and
c, is you can always

266
00:11:36,830 --> 00:11:39,340
just multiply through.

267
00:11:39,340 --> 00:11:40,990
And if you multiply through--

268
00:11:40,990 --> 00:11:41,310
OK.

269
00:11:41,310 --> 00:11:42,830
So we'll multiply through
on the left hand.

270
00:11:42,830 --> 00:11:45,850
We'll clear denominator, we'll
multiply through by x times 1

271
00:11:45,850 --> 00:11:46,500
plus x squared.

272
00:11:46,500 --> 00:11:49,240
So on the left we'll
get just 1.

273
00:11:49,240 --> 00:11:49,920
On the right--

274
00:11:49,920 --> 00:11:50,350
OK.

275
00:11:50,350 --> 00:11:56,945
So we multiply a over x times
x times 1 plus x squared, so

276
00:11:56,945 --> 00:11:59,780
that gives us a times 1 plus x
squared, but we know that a is

277
00:11:59,780 --> 00:12:05,220
1, so this is-- the first term
becomes 1 plus x squared, and

278
00:12:05,220 --> 00:12:08,560
the second term we multiply
through by x times 1 plus x

279
00:12:08,560 --> 00:12:10,920
squared, and the 1 plus x
squareds cancel, and we're

280
00:12:10,920 --> 00:12:18,850
left with plus x times
bx plus c.

281
00:12:18,850 --> 00:12:23,540
Now, if you, maybe you could
at this point, there are a

282
00:12:23,540 --> 00:12:24,810
number of different
ways to finish.

283
00:12:24,810 --> 00:12:27,230
But one, for example, is you
could see, all right, if you

284
00:12:27,230 --> 00:12:31,180
subtract the 1 plus x squared
over to the other side, you

285
00:12:31,180 --> 00:12:36,430
have minus x squared, and
over here, you have bx

286
00:12:36,430 --> 00:12:38,540
squared plus cx.

287
00:12:38,540 --> 00:12:41,850
So for those things to be
equal, you need b to be

288
00:12:41,850 --> 00:12:45,910
negative 1, and you
need c to be 0.

289
00:12:45,910 --> 00:12:47,490
Right?

290
00:12:47,490 --> 00:12:47,756
So OK.

291
00:12:47,756 --> 00:12:51,700
So good.

292
00:12:51,700 --> 00:12:58,240
So we have b equals minus
1, c equals 0.

293
00:12:58,240 --> 00:13:00,630
So the partial fraction
decomposition, then, if we

294
00:13:00,630 --> 00:13:12,300
plug this in, is 1 over x minus
x over 1 plus x squared.

295
00:13:12,300 --> 00:13:17,250
So this is the partial fraction
decomposition of this

296
00:13:17,250 --> 00:13:18,890
rational function.

297
00:13:18,890 --> 00:13:20,980
So you apply that partial
fraction decomposition.

298
00:13:20,980 --> 00:13:21,400
What do you do?

299
00:13:21,400 --> 00:13:24,230
All right, So now you carry
that back upstairs to the

300
00:13:24,230 --> 00:13:25,660
integral we were working on.

301
00:13:25,660 --> 00:13:29,170

302
00:13:29,170 --> 00:13:32,470
So our integral that we started
with is equal to,

303
00:13:32,470 --> 00:13:34,910
now-- well, so this constant
is still out in front.

304
00:13:34,910 --> 00:13:45,280
So it's minus 2 arc tan
x over x plus--

305
00:13:45,280 --> 00:13:47,110
so now we've got 2 times--

306
00:13:47,110 --> 00:13:47,370
all right.

307
00:13:47,370 --> 00:13:49,990
So first, this is just algebra
that we've done, so we can

308
00:13:49,990 --> 00:13:51,210
just substitute it in.

309
00:13:51,210 --> 00:13:56,440
So we replace this 1 over x
times 1 plus x squared by 1

310
00:13:56,440 --> 00:14:04,910
over x minus x over 1
plus x squared dx.

311
00:14:04,910 --> 00:14:05,410
OK.

312
00:14:05,410 --> 00:14:08,290
And now using the properties
of integration, this is a

313
00:14:08,290 --> 00:14:09,420
difference of two--

314
00:14:09,420 --> 00:14:11,180
an integral of a difference
is the

315
00:14:11,180 --> 00:14:13,440
difference of the two integrals.

316
00:14:13,440 --> 00:14:15,780
So we just integrate them
separately, right?

317
00:14:15,780 --> 00:14:17,440
So this is equal to--
so OK, the part

318
00:14:17,440 --> 00:14:19,460
out front never changes.

319
00:14:19,460 --> 00:14:26,330
2 arc tan x over x,
plus 2 times--

320
00:14:26,330 --> 00:14:26,580
OK.

321
00:14:26,580 --> 00:14:32,190
So you integrate 1 over x, and
that just gives you ln of

322
00:14:32,190 --> 00:14:34,180
absolute value of x.

323
00:14:34,180 --> 00:14:38,280
And OK, so you integrate, so
the second part is x over 1

324
00:14:38,280 --> 00:14:39,300
plus x squared.

325
00:14:39,300 --> 00:14:42,720
Now, in the worst case, we would
need to do some more

326
00:14:42,720 --> 00:14:44,420
work on this term and
split things up.

327
00:14:44,420 --> 00:14:45,820
And one of them, we might
have to complete

328
00:14:45,820 --> 00:14:46,820
the square or something.

329
00:14:46,820 --> 00:14:49,160
But in this case, it actually
has worked out kind of nicely.

330
00:14:49,160 --> 00:14:54,850
This x over 1 plus x squared is
a simple thing to handle.

331
00:14:54,850 --> 00:14:56,860
That's just ln of 1
plus x squared.

332
00:14:56,860 --> 00:15:00,410
Well, actually, 1/2 ln
of 1 plus x squared.

333
00:15:00,410 --> 00:15:09,500
So this is minus 1/2 ln
of 1 plus x squared.

334
00:15:09,500 --> 00:15:11,950
I could write absolute value,
but 1 plus x squared is always

335
00:15:11,950 --> 00:15:15,580
positive, so it doesn't
matter if I do or not.

336
00:15:15,580 --> 00:15:19,330
And if I wanted, I could
rewrite this a bit.

337
00:15:19,330 --> 00:15:22,290
I could rewrite this as-- just
to finish it, I could write 2

338
00:15:22,290 --> 00:15:30,780
arc tan x over x plus 2 ln
absolute value of x minus ln

339
00:15:30,780 --> 00:15:36,550
of 1 plus x squared.

340
00:15:36,550 --> 00:15:39,020
And I was doing an
antiderivative.

341
00:15:39,020 --> 00:15:40,930
So at some point, whenever I
finished the last one, I

342
00:15:40,930 --> 00:15:42,130
should have added a constant.

343
00:15:42,130 --> 00:15:45,350
So this should have had a plus
c here, and it should have a

344
00:15:45,350 --> 00:15:46,586
plus c there.

345
00:15:46,586 --> 00:15:47,836
Right, good.

346
00:15:47,836 --> 00:15:51,140

347
00:15:51,140 --> 00:15:54,350
So OK, so there's my answer.

348
00:15:54,350 --> 00:15:56,940
And let's just recap quickly
how we arrived at it.

349
00:15:56,940 --> 00:16:00,720
So we started with
this integral.

350
00:16:00,720 --> 00:16:04,150
And we saw, we recognized it
as a good candidate for

351
00:16:04,150 --> 00:16:06,260
integration by parts, and so
we applied integration by

352
00:16:06,260 --> 00:16:10,350
parts, seeing that the arc tan
part was the part that really

353
00:16:10,350 --> 00:16:13,500
wanted to be differentiated, and
that the 1 over x squared

354
00:16:13,500 --> 00:16:15,760
part, you know, it could have
been differentiated, it

355
00:16:15,760 --> 00:16:20,080
could've been integrated, but
since the arc tan wanted to be

356
00:16:20,080 --> 00:16:21,600
differentiated, this
got integrated.

357
00:16:21,600 --> 00:16:24,350

358
00:16:24,350 --> 00:16:24,560
Good.

359
00:16:24,560 --> 00:16:26,800
So we did integration by parts,
and then we're left

360
00:16:26,800 --> 00:16:30,250
with a rational function as the
piece of the integral we

361
00:16:30,250 --> 00:16:31,320
had left to compute.

362
00:16:31,320 --> 00:16:33,800
So when you have a rational
function, one method that

363
00:16:33,800 --> 00:16:36,690
always works is that you can
do partial fractions.

364
00:16:36,690 --> 00:16:38,890
Now, in this case, this was
a fairly simple partial

365
00:16:38,890 --> 00:16:39,910
fractions to do.

366
00:16:39,910 --> 00:16:44,260
First of all, the degree of the
numerator was smaller than

367
00:16:44,260 --> 00:16:46,550
the degree of the denominator,
so we didn't have to divide by

368
00:16:46,550 --> 00:16:48,920
anything, you know, we didn't
have to do long division.

369
00:16:48,920 --> 00:16:52,710
And the denominator had a fairly
simple form, so we

370
00:16:52,710 --> 00:16:55,650
could just do the cover up
method, and then multiply

371
00:16:55,650 --> 00:16:57,430
through and be done
fairly quickly.

372
00:16:57,430 --> 00:16:58,050
But in any case.

373
00:16:58,050 --> 00:17:01,410
So we did, we applied partial
fractions, so we got this

374
00:17:01,410 --> 00:17:02,110
expression.

375
00:17:02,110 --> 00:17:04,320
So then we carried that back
up to our integral, and

376
00:17:04,320 --> 00:17:05,950
integrated.

377
00:17:05,950 --> 00:17:08,690
OK, so then we had this integral
with the partial

378
00:17:08,690 --> 00:17:13,190
fraction expression in it, and
that was easy to integrate.

379
00:17:13,190 --> 00:17:16,540
So in this case, quite
easy to integrate.

380
00:17:16,540 --> 00:17:19,240
Sometimes it's a little messier,
but we came out

381
00:17:19,240 --> 00:17:20,586
pretty luckily this time.

382
00:17:20,586 --> 00:17:22,320
So OK.

383
00:17:22,320 --> 00:17:23,170
And so there you go.

384
00:17:23,170 --> 00:17:25,150
So that's going to be
our final answer.

385
00:17:25,150 --> 00:17:27,240
Now, this was long
and complicated.

386
00:17:27,240 --> 00:17:29,610
And so sometimes, when you
do a long and complicated

387
00:17:29,610 --> 00:17:31,530
computation, you worry,
did I make any stupid

388
00:17:31,530 --> 00:17:33,490
mistakes along the way?

389
00:17:33,490 --> 00:17:36,800
One way to check, always, when
you've done an antiderivative,

390
00:17:36,800 --> 00:17:38,990
is you could always check, you
could take the derivative of

391
00:17:38,990 --> 00:17:39,870
this expression.

392
00:17:39,870 --> 00:17:43,520
So that'll require you to use
some product rule and some

393
00:17:43,520 --> 00:17:45,240
chain rule.

394
00:17:45,240 --> 00:17:51,420
IBut, you know, arithmetically
a little difficult, but not

395
00:17:51,420 --> 00:17:52,760
sort of intellectually, right?

396
00:17:52,760 --> 00:17:56,470
You just are applying rules
sort of automatically.

397
00:17:56,470 --> 00:17:59,310
So you can take the derivative
of this expression, and check

398
00:17:59,310 --> 00:18:02,120
to make sure that it actually
agrees with the thing that we

399
00:18:02,120 --> 00:18:04,330
started with.

400
00:18:04,330 --> 00:18:05,820
So good.

401
00:18:05,820 --> 00:18:07,520
I'll end there.

402
00:18:07,520 --> 00:18:07,621