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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

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of different
integration techniques.

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So I have here a
couple of problems

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that will give
you an opportunity

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to apply those techniques, and
figure out which ones to apply.

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So in particular I have
one definite integral,

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the integral 0 to 2 of
the quantity x times e

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to the 1 minus x squared
with respect to x.

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And the second one is
an indefinite integral

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of 2 arctan x divided by x
squared with respect to x.

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

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to work these out, come back,
and we can work them out

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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

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go on to the second one.

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

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to the 1 minus x squared dx.

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

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

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off the top of my head?

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And this is not one
of the, you know,

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ones that we have
immediately already memorized

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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 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,

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the technique that is
going to work for you

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is just the simplest
technique that you have,

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which is just a
straightforward substitution.

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

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and you can see
that here we have

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a composition of functions,
this e to the 1 minus x

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squared, and so
whenever you have

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a composition of
functions, there's

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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

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to the 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,

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is that we're going to
put this inside function,

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this 1 minus x 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,

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u equal 1 minus x squared.

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So now if I take a differential,
if u equals 1 minus x squared,

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that means that
du is minus 2x dx.

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And since this is a
definite integral,

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I also need to 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

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that u is equal to 1 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--
so now the lower bound

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becomes u equals 1,
and the upper bound

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is u equals negative 3.

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OK.

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So now e to the 1
minus x squared,

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that's just e to the u.

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And x dx-- 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,

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a 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-- 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,

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we don't have 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-- all right,

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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

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to 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

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by substitution.

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Definite integral, change
bounds, and at the end

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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,

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and it 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 arctan 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

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that 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

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we could 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 arctan.

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I'm going to try and
get rid of arctan.

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And the way I'm going
to get rid of arctan

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is by doing a substitution,
u equals arctan x,

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or x equals tan u.

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So if you do that,
that'll get rid

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of your inverse trigonometric
function, but it'll introduce,

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in the bottom, you
have this x squared,

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so that'll introduce
like a tan u squared,

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and dx will give you
some more trig functions.

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

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this 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

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think that a polynomial,
or a power of u

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times a bunch of trig 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

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be 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,

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is that we 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 arctan
x times 1 over x squared.

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And arctan x 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

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behave like logarithms
with respect

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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 arctan,

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you get something 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

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and differentiate the arctan,
that makes your life nice.

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

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you need for everything
else to be something

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you can antidifferentiate, and
here everything else is just 2

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over x squared, and
so that's something

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that 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 going

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to set u equal to arctan 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,

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well, I guess I have to
pick up the 2 somewhere,

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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,

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so that 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,

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and that tells me this is
equal to-- well, let's see.

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So it's u*v, which is
minus 2 arctan x over x.

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That's u*v. Minus the integral
of, now it's the second part,

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it's v u prime dx.

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So here v u prime is-- 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-- 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,

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you have the u*v stuff that
gets kicked out out front,

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and then what you're left
with is this second integral.

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

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we 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

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this 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

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of this 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--
again, I have a 1

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plus x squared in
the denominator,

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and so that might
make you tempted to do

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a trig 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

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would be able 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--
well, on the one hand,

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this has this 1 plus x
squared in the denominator,

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but on the other hand, this
is just a 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

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on the 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

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always use-- what's it called--
partial fraction decomposition.

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So we can try to use
partial fractions

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on the second 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

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over x times 1 plus x squared.

228
00:10:01,722 --> 00:10:03,680
So if you remember your
partial fractions here,

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this is completely factored.

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This quadratic doesn't factor.

231
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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,

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and a single non-factorable
quadratic term.

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So partial fractions tells
us that we can write this

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

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quadratic, it needs to be B*x
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,

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you get two 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

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of the denominator
is 3, so that's

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one way of checking that
you're not completely off base,

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that the degree down
here should always

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agree with the
number of constants

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that you're solving for.

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OK.

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00:10:53,380 --> 00:10:53,880
Good.

249
00:10:53,880 --> 00:10:57,310
So now we need to solve
this for A, B, and C.

250
00:10:57,310 --> 00:11:00,620
And so OK, because we have
this nice single linear factor,

251
00:11:00,620 --> 00:11:02,730
we can do the cover
up method there.

252
00:11:02,730 --> 00:11:07,776
So we cover up x, and
we cover up over x,

253
00:11:07,776 --> 00:11:09,384
and we cover up everything else.

254
00:11:09,384 --> 00:11:10,800
So on the right-hand
side, we just

255
00:11:10,800 --> 00:11:14,290
end up with A, and on the
left-hand side, well, so x.

256
00:11:14,290 --> 00:11:16,850
So that's what we need, we
get whatever when we plug in

257
00:11:16,850 --> 00:11:17,960
x equals 0.

258
00:11:17,960 --> 00:11:21,900
So on this side, we get
1 over 1 plus 0 squared,

259
00:11:21,900 --> 00:11:25,060
so that's just 1.

260
00:11:25,060 --> 00:11:29,770
So the cover up method gives
us that A is equal to 1.

261
00:11:29,770 --> 00:11:31,630
And now we need to
solve for B and C.

262
00:11:31,630 --> 00:11:33,590
And probably the most
straightforward way,

263
00:11:33,590 --> 00:11:35,190
in this case, to
solve for B and C,

264
00:11:35,190 --> 00:11:39,340
is you can always
just multiply through.

265
00:11:39,340 --> 00:11:40,997
And if you multiply
through-- OK.

266
00:11:40,997 --> 00:11:42,830
So we'll multiply through
on the left-hand--

267
00:11:42,830 --> 00:11:44,830
we'll clear denominators,
we'll multiply through

268
00:11:44,830 --> 00:11:46,500
by x times 1 plus x squared.

269
00:11:46,500 --> 00:11:49,240
So on the left we'll get just 1.

270
00:11:49,240 --> 00:11:50,350
On the right-- OK.

271
00:11:50,350 --> 00:11:56,830
So we multiply A over x times
x times 1 plus x squared,

272
00:11:56,830 --> 00:11:58,890
so that gives us A
times 1 plus x squared,

273
00:11:58,890 --> 00:12:00,990
but we know that A
is 1, so this is--

274
00:12:00,990 --> 00:12:05,080
the first term becomes
1 plus x squared,

275
00:12:05,080 --> 00:12:08,560
and the second term we multiply
through by x times 1 plus x

276
00:12:08,560 --> 00:12:10,700
squared, and the 1
plus x squareds cancel,

277
00:12:10,700 --> 00:12:18,850
and we're left with
plus x times B*x plus C.

278
00:12:18,850 --> 00:12:22,820
Now, if you, maybe you
could-- at this point,

279
00:12:22,820 --> 00:12:24,810
there are a number of
different ways to finish.

280
00:12:24,810 --> 00:12:27,040
But one, for example, is
you could see, all right,

281
00:12:27,040 --> 00:12:30,970
if you subtract the 1 plus x
squared over to the other side,

282
00:12:30,970 --> 00:12:35,230
you have minus x
squared, and over here,

283
00:12:35,230 --> 00:12:38,540
you have B x squared plus C*x.

284
00:12:38,540 --> 00:12:43,040
So for those things to be equal,
you need B to be negative 1,

285
00:12:43,040 --> 00:12:45,910
and you need C to be 0.

286
00:12:45,910 --> 00:12:47,256
Right?

287
00:12:47,256 --> 00:12:47,756
So OK.

288
00:12:47,756 --> 00:12:51,700
So good.

289
00:12:51,700 --> 00:12:57,817
So we have B equals
minus 1, C equals 0.

290
00:12:57,817 --> 00:12:59,400
So the partial
fraction decomposition,

291
00:12:59,400 --> 00:13:07,830
then, if we plug this in,
is 1 over x minus x over 1

292
00:13:07,830 --> 00:13:12,300
plus x squared.

293
00:13:12,300 --> 00:13:14,720
So this is the partial
fraction decomposition

294
00:13:14,720 --> 00:13:18,890
of this rational function.

295
00:13:18,890 --> 00:13:20,980
So you apply that partial
fraction decomposition.

296
00:13:20,980 --> 00:13:21,605
What do you do?

297
00:13:21,605 --> 00:13:24,640
All right, So now you carry that
back upstairs to the integral

298
00:13:24,640 --> 00:13:25,660
we were working on.

299
00:13:29,170 --> 00:13:31,700
So our integral
that we started with

300
00:13:31,700 --> 00:13:34,910
is equal to, now-- well, so this
constant is still out in front.

301
00:13:34,910 --> 00:13:45,420
So it's minus 2 arctan
x over x plus-- so

302
00:13:45,420 --> 00:13:47,370
now we've got 2
times-- all right.

303
00:13:47,370 --> 00:13:49,700
So first, this is just
algebra that we've done,

304
00:13:49,700 --> 00:13:51,210
so we can just substitute it in.

305
00:13:51,210 --> 00:13:54,200
So we replace this
1 over x times 1

306
00:13:54,200 --> 00:14:04,910
plus x squared by 1 over x minus
x over 1 plus x squared dx.

307
00:14:04,910 --> 00:14:05,410
OK.

308
00:14:05,410 --> 00:14:07,810
And now using the
properties of integration,

309
00:14:07,810 --> 00:14:10,510
this is a difference of two--
an integral of a difference

310
00:14:10,510 --> 00:14:13,440
is the difference of
the two integrals.

311
00:14:13,440 --> 00:14:15,780
So we just integrate
them separately, right?

312
00:14:15,780 --> 00:14:19,460
So this is equal to-- so OK, the
part out front never changes.

313
00:14:19,460 --> 00:14:26,580
2 arctan x over x,
plus 2 times-- OK.

314
00:14:26,580 --> 00:14:30,200
So you integrate 1 over
x, and that just gives you

315
00:14:30,200 --> 00:14:34,180
ln of absolute value of x.

316
00:14:34,180 --> 00:14:38,280
And OK, so you integrate-- so
the second part is x over 1

317
00:14:38,280 --> 00:14:39,300
plus x squared.

318
00:14:39,300 --> 00:14:41,750
Now, in the worst
case, we would need

319
00:14:41,750 --> 00:14:44,062
to do some more work on this
term and split things up.

320
00:14:44,062 --> 00:14:46,270
And one of them, we might
have to complete the square

321
00:14:46,270 --> 00:14:46,820
or something.

322
00:14:46,820 --> 00:14:49,320
But in this case, it actually
has worked out kind of nicely.

323
00:14:49,320 --> 00:14:54,850
This x over 1 plus x squared
is a simple thing to handle.

324
00:14:54,850 --> 00:14:56,860
That's just ln of
1 plus x squared.

325
00:14:56,860 --> 00:15:00,410
Well, actually, one half
of ln of 1 plus x squared.

326
00:15:00,410 --> 00:15:09,500
So this is minus 1/2
ln of 1 plus x squared.

327
00:15:09,500 --> 00:15:11,640
I could write absolute
value, but 1 plus x squared

328
00:15:11,640 --> 00:15:15,517
is always positive, so it
doesn't matter if I do or not.

329
00:15:15,517 --> 00:15:17,350
And if I wanted, I could
rewrite this a bit.

330
00:15:17,350 --> 00:15:20,356
I could rewrite this
as-- just to finish it,

331
00:15:20,356 --> 00:15:27,200
I could write 2
arctan x over x plus 2

332
00:15:27,200 --> 00:15:36,550
ln absolute value of x minus
ln of 1 plus x squared.

333
00:15:36,550 --> 00:15:38,746
And I was doing
an antiderivative.

334
00:15:38,746 --> 00:15:40,870
So at some point, whenever
I finished the last one,

335
00:15:40,870 --> 00:15:42,161
I should have added a constant.

336
00:15:42,161 --> 00:15:44,440
So this should have
had a plus c here,

337
00:15:44,440 --> 00:15:48,050
and it should have
a plus c there.

338
00:15:48,050 --> 00:15:51,140
Right, good.

339
00:15:51,140 --> 00:15:54,350
So OK, so there's my answer.

340
00:15:54,350 --> 00:15:56,940
And let's just recap quickly
how we arrived at it.

341
00:15:56,940 --> 00:16:00,720
So we started with
this integral.

342
00:16:00,720 --> 00:16:03,859
And we saw, we recognized
it as a good candidate

343
00:16:03,859 --> 00:16:06,150
for integration by parts,
and so we applied integration

344
00:16:06,150 --> 00:16:10,350
by parts, seeing that the arctan
part was the part that really

345
00:16:10,350 --> 00:16:13,837
wanted to be differentiated, and
that the 1 over x squared part,

346
00:16:13,837 --> 00:16:15,670
you know, it could have
been differentiated,

347
00:16:15,670 --> 00:16:18,170
it could've been integrated,
but since the arctan

348
00:16:18,170 --> 00:16:21,600
wanted to be differentiated,
this got integrated.

349
00:16:24,060 --> 00:16:24,560
Good.

350
00:16:24,560 --> 00:16:26,520
So we did integration
by parts, and then we're

351
00:16:26,520 --> 00:16:30,250
left with a rational function
as the piece of the integral we

352
00:16:30,250 --> 00:16:31,320
had left to compute.

353
00:16:31,320 --> 00:16:32,880
So when you have a
rational function,

354
00:16:32,880 --> 00:16:35,140
one method that always
works is that you

355
00:16:35,140 --> 00:16:36,690
can do partial fractions.

356
00:16:36,690 --> 00:16:39,380
Now, in this case, this was a
fairly simple partial fractions

357
00:16:39,380 --> 00:16:39,910
to do.

358
00:16:39,910 --> 00:16:42,930
First of all, the
degree of the numerator

359
00:16:42,930 --> 00:16:45,400
was smaller than the
degree of the denominator,

360
00:16:45,400 --> 00:16:47,320
so we didn't have to
divide by anything,

361
00:16:47,320 --> 00:16:49,194
you know, we didn't have
to do long division.

362
00:16:49,194 --> 00:16:52,244
And the denominator had
a fairly simple form,

363
00:16:52,244 --> 00:16:53,910
so we could just do
the cover up method,

364
00:16:53,910 --> 00:16:57,384
and then multiply through
and be done fairly quickly.

365
00:16:57,384 --> 00:16:58,050
But in any case.

366
00:16:58,050 --> 00:17:00,550
So we did, we applied
partial fractions,

367
00:17:00,550 --> 00:17:02,110
so we got this expression.

368
00:17:02,110 --> 00:17:04,180
So then we carried that
back up to our integral,

369
00:17:04,180 --> 00:17:05,950
and integrated.

370
00:17:05,950 --> 00:17:09,330
OK, so then we had this integral
with the partial fraction

371
00:17:09,330 --> 00:17:13,190
expression in it, and that
was easy to integrate.

372
00:17:13,190 --> 00:17:16,540
So in this case, quite
easy to integrate.

373
00:17:16,540 --> 00:17:19,240
Sometimes it's a little
messier, but we came out

374
00:17:19,240 --> 00:17:20,586
pretty luckily this time.

375
00:17:20,586 --> 00:17:22,320
So OK.

376
00:17:22,320 --> 00:17:23,170
And so there you go.

377
00:17:23,170 --> 00:17:25,150
So that's going to
be our final answer.

378
00:17:25,150 --> 00:17:27,240
Now, this was long
and complicated.

379
00:17:27,240 --> 00:17:29,610
And so sometimes, when you
do a long and complicated

380
00:17:29,610 --> 00:17:31,890
computation, you worry, did
I make any stupid mistakes

381
00:17:31,890 --> 00:17:33,490
along the way?

382
00:17:33,490 --> 00:17:36,447
One way to check, always, when
you've done an antiderivative,

383
00:17:36,447 --> 00:17:38,780
is you could always check,
you could take the derivative

384
00:17:38,780 --> 00:17:39,870
of this expression.

385
00:17:39,870 --> 00:17:42,070
So that'll require you
to use some product

386
00:17:42,070 --> 00:17:44,460
rule and some chain rule.

387
00:17:44,460 --> 00:17:48,320
But, you know,
it's arithmetically

388
00:17:48,320 --> 00:17:52,760
a little difficult, but not
sort of intellectually, right?

389
00:17:52,760 --> 00:17:56,470
You just are applying rules
sort of automatically.

390
00:17:56,470 --> 00:17:58,860
So you can take the
derivative of this expression,

391
00:17:58,860 --> 00:18:01,940
and check to make sure that it
actually agrees with the thing

392
00:18:01,940 --> 00:18:04,330
that we started with.

393
00:18:04,330 --> 00:18:05,820
So good.

394
00:18:05,820 --> 00:18:07,121
I'll end there.