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Welcome back to recitation.

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In this video, we're going to be
working on establishing the

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best technique for finding
an integral or finding an

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antiderivative.

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We'll be doing this, as you've
seen probably in a lot of

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these videos, in a row.

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And so in this one in
particular, we're going to

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work on these two.

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So what I'd like us to find, is
for the letter A, I'd like

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us to find an actual value if
we take the integral from

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minus 1 to 0 of this fraction.

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5x squared minus 2x plus 3 over
the quantity x squared

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plus 1 times x minus one.

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And then, the second problem,
we're just going to be finding

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an antiderivative.

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So it's finding an
antiderivative of the function

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1 over x plus 1 times the square
root of negative x

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squared minus 2x.

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Now, this does have a domain
over which this function is

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well-defined, as long as what's
inside the square root

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is positive.

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And there are values of x for
which that's positive.

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I just didn't want us to have
to compute this one exactly.

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So we're just looking for
an antiderivative here.

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So the goal is, figure out what
strategy you want to use,

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work through that strategy,
and then I'll be back, and

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I'll show you which strategy
I picked, and the

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solution that I got.

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OK.

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Welcome back.

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Well, hopefully you were
able to make some

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headway in both of these.

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And so what we'll do right away,
is just we'll start with

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the first one.

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So on the first one, we should
be able to get an actual

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numerical value at the
conclusion of the problem.

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And if you look at the first
one, it's probably pretty

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obvious you want to use
partial fraction

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decomposition.

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You already have a denominator
that's factored, and so this

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is going to be fairly
easy to do.

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Now, what's one thing you want
to check with partial

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fractions, is you want to make
sure that the degree of the

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numerator is smaller than the
degree of the denominator.

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Notice that the numerator
degree is 2 and the

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denominator degree is 3,
because we have an x

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squared times x.

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So we don't have to do
any long division.

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We can just start the problem.

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So what I'm going to do, is
I'm going to actually

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decompose it without showing
you how I did it.

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And you've done that practice
enough, so I'm just going to

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show you what I got with my
decomposition, and we'll go

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from there.

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So when I decompose
for letter A, I

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actually get two integrals.

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And the first one I get is the
integral from minus 1 to 0 of

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2x over x squared plus 1 dx.

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And the second one I get is the
integral minus 1 to 0 of 3

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over x minus 1 dx.

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Let me just double check
and make sure-- yes.

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That's what I got when I did
this problem earlier.

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So this is not magic.

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I actually did this already.

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That's how I got these.

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And now from here, we just
have to determine--

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we just have to integrate
both of these.

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Now, what would be the strategy
at this point?

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Well this one--

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if this had just been a 2 and no
x, you'd be dealing with an

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arc tan type of problem.

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Because you'd be integrating
1 over x squared plus 1.

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But actually, because we have
this 2x here, this is really a

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substitution problem.

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Right?

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The derivative of x squared
plus 1 is 2x.

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Right?

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So we see that really what we're
integrating is something

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like du over u.

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And so if you did this
substitution problem, you

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should get something like
natural log of x

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squared plus 1.

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Let me just double check and
make sure that gives me the

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derivative here.

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The derivative of natural log of
x squared plus 1 is 1 over

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x squared plus 1 times 2x.

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That gives me exactly
what's here.

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I don't need absolute
values here, because

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this is always positive.

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And I know I need to evaluate
it at minus 1 and 0.

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OK?

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So that takes care
of the left one.

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Now, the right one is again,
it's a pretty straightforward

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one, because it's just
3 over x minus 1.

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That's a natural log again.

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So this natural log
is even simpler.

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It's going to be 3 times the
natural log absolute x minus 1

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from minus 1 to 0.

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And so now we just have
to plug in everything.

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OK?

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So let's just do this one step
at a time, starting over here.

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So the natural log, when
I put in 0, I get the

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natural log of 1.

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That's 0.

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And I subtract what I get when
I put in negative 1 for x.

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And negative 1 squared gives
me 1, so this is minus the

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natural log of 2.

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And then I have plus 3 times
whatever's over here.

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So now let's look at this.

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When I plug in 0, I get natural
log of 0 minus 1

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absolute value.

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That's natural log of 1.

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That's zero again.

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And then I get a minus.

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And then I put in negative 1.

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Negative 1 minus 1, negative
2, absolute value, so it's

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natural log of 2.

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And so if I look it at all the
way across, I see I have a

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negative natural log of
2 and then I have 3

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natural logs of 2.

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So the final answer is just
negative 4 natural log of 2.

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And that is where we'll

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stop with A. OK.

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So let me just remind you,
actually, before we go to B.

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What we did in A was we
did partial fraction

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decomposition.

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And I gave you the numerators.

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And then on the first one,
we had to use maybe a

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substitution to figure it out.

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I didn't write explicitly
the substitution, but a

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substitution gives us that
integral, and this one is

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directly a natural log.

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OK.

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Now let's look at B. So B--

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let me rewrite the problem,
because it's now

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a little far away.

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I think it's x plus 1 square
root of negative x

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squared minus 2x.

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OK.

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So B, the reason--

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I wanted to make sure we did
a trig substitution in a

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particular way, because
I haven't

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demonstrated those very much.

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So the denominator wound up
looking a little awkward, to

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force you to do it
in that way.

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But what we want to do, is we
want to actually complete the

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square on what's in here.

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And that might make you
a little bit nervous.

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But let me just do a little
sidebar work down here, and

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point out what we get.

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If we factor out a negative
here, we get an x

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squared plus 2x.

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OK?

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So we're going to complete
the square on the inside.

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Now this might make some
people nervous.

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They might say, you've
a negative

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under the square root.

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But I want to point out that I
have a negative here, but I

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could always make x squared plus
2x a negative number, and

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then I would have--

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the negative times a negative
is a positive.

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For instance, I think negative
1, right, if I put a negative

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1 for x, I get negative 2 plus
1 is a negative value.

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So if I put a negative 1 for x,
I'm taking the square root

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of a positive number.

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So there are values of x that
make this under the square

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root positive.

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OK?

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So you don't have to
worry about that.

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Now, if I want to complete
the square on what's in

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here, what do I do?

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I have x squared plus 2x.

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I obviously need to add a 1
to complete the square.

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Why is that?

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Because you take what's here,
you divide it by 2,

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and you square it.

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And so this actually equals
square root of negative x

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squared plus 2x plus 1.

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And I want to subtract 1 so
that I haven't changed

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anything, but when I pull it
out from the negative, it's

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another plus 1.

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OK?

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Let's make sure we buy that.

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I've added 1 inside here, so
if I add 1 on the outside,

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this is actually a minus 1, and
so this is a plus 1, so

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together they add up to 0.

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So I haven't changed what's
in the square root.

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So if I come back and put that
in right here, what do I get?

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I get the integral dx over x
plus 1 square root-- let me

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move this over.

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I'm going to bring this
to the front--

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1 minus x plus 1 squared.

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All right.

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So from here, we have to
do a trig substitution.

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Now, what trig substitution
we want to do, we

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can do sine or cosine.

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But I'm going to do cosine,
because I like secants better

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than cosecants, because I have
those memorized better.

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So that's why I'm
choosing cosine.

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You'll see why I chose that
way in a little bit.

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But it would be, you
will get the same

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answer if you do sine.

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OK.

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So I'm going to come
to the other side.

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Let's see.

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So I'm choosing cosine theta
equals x plus 1.

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That's the substitution
I'm making.

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And why am I making
that substitution?

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I'm making that substitution
because when I do 1 minus x

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plus 1 squared, that's actually
1 minus cosine

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squared theta.

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So that's, this in here
is sine squared theta.

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And when I take the square root,
I just get a sine theta.

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So that should be pretty
familiar to you

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by now, this strategy.

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But the point I'm making is that
x plus 1 will be a cosine

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00:09:02,800 --> 00:09:05,180
theta, and this whole
square root is what

222
00:09:05,180 --> 00:09:06,300
becomes a sine theta.

223
00:09:06,300 --> 00:09:08,050
So you've seen that a fair
amount, but just

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to remind you that.

225
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And then the other thing we
need is to replace the dx.

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So the dx is going to be,
derivative of cosine is

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negative sine, so you're
going to get negative

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sine theta d theta.

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00:09:21,860 --> 00:09:23,040
So now we know all the pieces.

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We said this was cosine, we said
the square root is sine,

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and the dx is negative
sine d theta.

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So let's rewrite
that over here.

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00:09:31,285 --> 00:09:34,180

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So we have--

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I'm going to put the negative
in front, so I don't have to

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deal with it anymore.

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Negative sine theta
over cosine theta

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sine theta d theta.

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These divide out, and I get
negative 1 over cosine theta,

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which is just equal to negative
secant theta.

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OK?

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So I have negative
secant theta.

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Let me actually write
that here.

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Negative integral of secant
theta d theta.

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And now what is that?

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00:10:02,200 --> 00:10:05,820
Well, we know how to
integrate secant.

247
00:10:05,820 --> 00:10:09,180
So let me write that
in terms of theta.

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00:10:09,180 --> 00:10:13,990
It's going to be negative
natural log absolute value

249
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secant theta plus
tangent theta.

250
00:10:17,670 --> 00:10:19,800
And then we have the
plus c out here.

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What's the point of this?

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Well, we should maybe
have this memorized.

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If you have to look it up, you
have to look it up, but you

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00:10:25,690 --> 00:10:27,080
saw this one in class.

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00:10:27,080 --> 00:10:29,560
And the negative is just
dropping down here, so don't

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00:10:29,560 --> 00:10:31,610
think I added that negative
in when I was taking

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00:10:31,610 --> 00:10:32,360
antiderivative.

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00:10:32,360 --> 00:10:33,740
It was already there.

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00:10:33,740 --> 00:10:34,070
All right.

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00:10:34,070 --> 00:10:34,930
So we're done.

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00:10:34,930 --> 00:10:36,130
Oh, but we're not done.

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00:10:36,130 --> 00:10:37,940
Why are we not done?

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00:10:37,940 --> 00:10:40,580
We're not done, because we
started off with something in

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00:10:40,580 --> 00:10:42,850
terms of x, and now we have
something in terms of theta,

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00:10:42,850 --> 00:10:44,330
so we have to finish up.

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00:10:44,330 --> 00:10:46,450
And how we do that, is we
go back, we look at the

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00:10:46,450 --> 00:10:47,520
substitution we made.

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00:10:47,520 --> 00:10:50,870
If we make a triangle based on
that substitution, we figure

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00:10:50,870 --> 00:10:53,760
out the values of secant theta
and tangent theta, and then we

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00:10:53,760 --> 00:10:56,340
can plug those in terms of x.

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00:10:56,340 --> 00:10:57,220
So let's remind ourselves--

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00:10:57,220 --> 00:10:59,060
I'm going to draw the triangle
in the middle here.

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00:10:59,060 --> 00:11:02,290
Let's remind ourselves of the
relationship we had between

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00:11:02,290 --> 00:11:05,100
theta and x.

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00:11:05,100 --> 00:11:10,270
If this is theta, we said cosine
theta, right here,

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00:11:10,270 --> 00:11:12,870
cosine theta was equal
to x plus 1.

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00:11:12,870 --> 00:11:16,300
Cosine theta is adjacent
over hypotenuse.

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00:11:16,300 --> 00:11:20,500
So we want to say, this is
x plus 1, and this is 1.

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00:11:20,500 --> 00:11:23,240
And that implies by the
Pythagorean theorem, that this

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00:11:23,240 --> 00:11:27,856
is square root of 1 minus
quantity x plus 1 squared.

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00:11:27,856 --> 00:11:29,790
Let me move that over.

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00:11:29,790 --> 00:11:32,290
Notice, then, this also
makes sense, why sine

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00:11:32,290 --> 00:11:33,300
theta is what it is.

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00:11:33,300 --> 00:11:36,000
Sine theta is this value
divided by 1.

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00:11:36,000 --> 00:11:38,650
So that also helps you
understand that.

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00:11:38,650 --> 00:11:39,100
All right.

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00:11:39,100 --> 00:11:40,480
So now what do we need
to read off?

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00:11:40,480 --> 00:11:41,850
We need to read off
secant, and we

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00:11:41,850 --> 00:11:43,390
need to read off tangent.

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00:11:43,390 --> 00:11:46,240
So secant is 1 over cosine, so
actually, we could have gotten

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00:11:46,240 --> 00:11:48,640
that one for free,
from the cosine.

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00:11:48,640 --> 00:11:52,730
So this 1 over cosine
is 1 over x plus 1.

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00:11:52,730 --> 00:11:56,650
So this thing is equal to
negative natural log absolute

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00:11:56,650 --> 00:12:00,110
value 1 over x plus 1 plus--

295
00:12:00,110 --> 00:12:01,590
now, what's tangent?

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00:12:01,590 --> 00:12:04,480
If I come back and look at the
triangle, tangent theta is

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00:12:04,480 --> 00:12:06,910
opposite over adjacent.

298
00:12:06,910 --> 00:12:07,460
Right?

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00:12:07,460 --> 00:12:11,380
So I can actually just put it
all over x plus 1 if I wanted.

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00:12:11,380 --> 00:12:13,760
But I already started writing
it separately, so I'll leave

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00:12:13,760 --> 00:12:15,090
it like this.

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00:12:15,090 --> 00:12:20,380
Square root of 1 minus x plus
1 quantity squared.

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00:12:20,380 --> 00:12:23,910
And then close that,
and then my plus c.

304
00:12:23,910 --> 00:12:26,520
So now I'm actually finished
with the problem.

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00:12:26,520 --> 00:12:31,220
Because now I have an
antiderivative in terms of x.

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00:12:31,220 --> 00:12:33,540
So let me just remind you where
this problem, where we

307
00:12:33,540 --> 00:12:36,330
started the problem, kind of
take us through quickly, and

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00:12:36,330 --> 00:12:37,340
then we'll be done.

309
00:12:37,340 --> 00:12:42,400
So back to the beginning, what
we had, was we had an integral

310
00:12:42,400 --> 00:12:46,460
that was a fractional problem,
but we had an x plus 1 here,

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00:12:46,460 --> 00:12:48,240
and then we had this really
messy-looking

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00:12:48,240 --> 00:12:49,850
quadratic in here.

313
00:12:49,850 --> 00:12:52,160
To make it easy to deal with,
I factored out a negative

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00:12:52,160 --> 00:12:54,880
sign, and then I saw I could
complete the square.

315
00:12:54,880 --> 00:12:58,270
Once you complete the square,
you actually get another x

316
00:12:58,270 --> 00:13:00,930
plus 1 in there, which helps
us to see immediately, it

317
00:13:00,930 --> 00:13:03,080
should be a trig substitution.

318
00:13:03,080 --> 00:13:05,410
So the substitution that's
natural to make, because you

319
00:13:05,410 --> 00:13:08,430
have a 1 minus something
involving an x, is going to be

320
00:13:08,430 --> 00:13:09,980
either cosine or sine.

321
00:13:09,980 --> 00:13:11,810
I chose cosine.

322
00:13:11,810 --> 00:13:13,700
If you'd chosen sine, you
probably would have gotten a

323
00:13:13,700 --> 00:13:17,715
cosecant up there, instead of a
secant, when you were taking

324
00:13:17,715 --> 00:13:19,470
an antiderivative
at the very end.

325
00:13:19,470 --> 00:13:22,450
So you would have gotten the
same answer because of the

326
00:13:22,450 --> 00:13:25,440
substitutions in the end.

327
00:13:25,440 --> 00:13:28,370
But so I chose cosine theta
is equal to x plus 1.

328
00:13:28,370 --> 00:13:31,000
You do that, you can replace
this with cosine, this with a

329
00:13:31,000 --> 00:13:34,700
sine, this becomes a negative
sine, and then you start

330
00:13:34,700 --> 00:13:35,450
simplifying.

331
00:13:35,450 --> 00:13:38,640
So once we came over here and
simplified, we got it into

332
00:13:38,640 --> 00:13:39,720
something we recognize.

333
00:13:39,720 --> 00:13:41,140
We got it into secant.

334
00:13:41,140 --> 00:13:45,085
We know the antiderivative
for secant, in terms

335
00:13:45,085 --> 00:13:46,110
of secant and tangent.

336
00:13:46,110 --> 00:13:47,730
We know it's exactly this.

337
00:13:47,730 --> 00:13:50,730
And then we went back to the
relationship we had.

338
00:13:50,730 --> 00:13:53,680
We made ourselves a triangle
in terms of the

339
00:13:53,680 --> 00:13:55,130
theta and the x-values.

340
00:13:55,130 --> 00:13:57,130
And then we were able
to substitute in

341
00:13:57,130 --> 00:13:59,300
for secant and tangent.

342
00:13:59,300 --> 00:14:00,040
All right.

343
00:14:00,040 --> 00:14:02,140
So hopefully that was
successful for you.

344
00:14:02,140 --> 00:14:04,460
And that's where I'll stop.

345
00:14:04,460 --> 00:14:04,645