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Welcome back to recitation.

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In this video, we're
going to be working

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on establishing the best
technique for finding

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an integral or finding
an antiderivative.

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We'll be doing this,
as you've seen probably

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in a lot of these
videos, in a row.

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And so in this
one in particular,

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we're going to
work on these two.

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

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I'd like us to find
an actual value

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if we take the integral from
minus 1 to 0 of this fraction.

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5 x squared minus 2x plus 3 over
the quantity x squared plus 1

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times x minus one.

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And then, the second
problem, we're

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just going to be finding
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 squared

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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,

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and I'll show you which
strategy I picked,

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and the 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 headway

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in both of these.

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

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we'll start with the first one.

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So on the first
one, we should be

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able to get an actual numerical
value at the conclusion

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of the problem.

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And if you look
at the first one,

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it's probably pretty obvious
you want to use partial fraction

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decomposition.

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You already have a
denominator that's factored,

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and so this is going to
be fairly easy to do.

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Now, what's one thing
you want to check

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with partial
fractions, is you want

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to make sure that the
degree of the numerator

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is smaller than the
degree of the denominator.

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Notice that the
numerator degree is 2

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and the denominator
degree is 3, because we

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have an x 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 decompose it

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without showing
you how I did it.

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And you've done that
practice enough,

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so I'm just going
to show you what

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I got with my decomposition,
and we'll go from there.

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So when I decompose,
for letter A,

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I actually get two integrals.

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And the first one I get is
the integral from minus 1

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to 0 of 2x over x
squared plus 1 dx.

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

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of 3 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-- we just have

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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-- if this
had just been a 2 and no x,

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you'd be dealing with an
arctan 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,

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this is really a
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

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is something like du over u.

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And so if you did this
substitution problem,

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you should get something
like natural log

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of x squared plus 1.

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Let me just double
check and make sure that

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gives me the derivative here.

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The derivative of natural
log of x squared plus 1

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is 1 over 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,

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because 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

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a pretty straightforward
one, because it's just 3

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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,

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I get the 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,

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so this is minus the
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

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minus 1, 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,

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so it's natural log of 2.

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

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I see I have a negative
natural log of 2

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and then I have 3
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 stop with (a).

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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 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 substitution

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to figure it out.

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I didn't write explicitly
the substitution,

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but a substitution
gives us that integral,

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and this one is
directly a natural log.

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OK.

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Now let's look at (b).

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So (b)-- let me rewrite
the problem, because it's

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now a little far away.

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

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minus 2x.

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OK.

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So (b), the reason--
I wanted to make sure

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we did a trig substitution
in a particular way,

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because I haven't
demonstrated those very much.

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So the denominator wound up
looking a little awkward,

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to force you to
do it in that way.

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

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the 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,

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and point out what we get.

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If we factor out
a negative here,

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we get an x 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 under the square root.

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

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but I could always
make x squared

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plus 2x a negative
number, and then

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I would have-- the negative
times a negative is a positive.

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For instance, I think
negative 1, right,

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if I put a negative 1 for
x, I get negative 2 plus 1

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is a negative value.

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So if I put a
negative 1 for x, I'm

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taking the square root
of a positive number.

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

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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 here,

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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

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of negative x squared
plus 2x plus 1.

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

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

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so 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--

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let me move this over.

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I'm going to bring
this to the front-- 1

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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,

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we can do sine or cosine.

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

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better 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 answer if you do sine.

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OK.

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00:08:28,870 --> 00:08:31,390
So I'm going to come
to the other side.

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Let's see.

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00:08:32,120 --> 00:08:37,470
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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00:08:39,650 --> 00:08:41,620
And why am I making
that substitution?

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00:08:41,620 --> 00:08:43,820
I'm making that
substitution because when

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I do 1 minus x plus 1
squared, that's actually

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00:08:47,140 --> 00:08:50,540
1 minus cosine squared theta.

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00:08:50,540 --> 00:08:53,840
So that's, this in here
is sine squared theta.

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00:08:53,840 --> 00:08:56,644
And when I take the square
root, I just get a sine theta.

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00:08:56,644 --> 00:08:58,310
So that should be
pretty familiar to you

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00:08:58,310 --> 00:08:59,800
by now, this strategy.

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But the point I'm
making is that x plus 1

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00:09:02,010 --> 00:09:04,920
will be a cosine theta,
and this whole square root

216
00:09:04,920 --> 00:09:06,300
is what becomes a sine theta.

217
00:09:06,300 --> 00:09:07,740
So you've seen
that a fair amount,

218
00:09:07,740 --> 00:09:09,770
but just to remind you that.

219
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And then the other thing we
need is to replace the dx.

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00:09:13,000 --> 00:09:16,471
So the dx is going to
be, derivative of cosine

221
00:09:16,471 --> 00:09:18,720
is negative sine, so you're
going to get negative sine

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00:09:18,720 --> 00:09:21,790
theta d theta.

223
00:09:21,790 --> 00:09:23,040
So now we know all the pieces.

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00:09:23,040 --> 00:09:26,530
We said this was cosine, we
said the square root is sine,

225
00:09:26,530 --> 00:09:29,440
and the dx is
negative sine d theta.

226
00:09:29,440 --> 00:09:31,285
So let's rewrite that over here.

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So we have-- I'm going to
put the negative in front,

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00:09:36,740 --> 00:09:38,640
so I don't have to
deal with it anymore.

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00:09:38,640 --> 00:09:47,030
Negative sine theta over cosine
theta sine theta d theta.

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00:09:47,030 --> 00:09:49,240
These divide out,
and I get negative 1

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00:09:49,240 --> 00:09:53,330
over cosine theta, which is just
equal to negative secant theta.

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00:09:53,330 --> 00:09:54,320
OK?

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00:09:54,320 --> 00:09:55,727
So I have negative secant theta.

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00:09:55,727 --> 00:09:57,060
Let me actually write that here.

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00:09:57,060 --> 00:10:00,940
Negative integral of
secant theta d theta.

236
00:10:00,940 --> 00:10:02,200
And now what is that?

237
00:10:02,200 --> 00:10:05,820
Well, we know how
to integrate secant.

238
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

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00:10:13,990 --> 00:10:17,670
secant theta plus tangent theta.

241
00:10:17,670 --> 00:10:19,800
And then we have
the plus c out here.

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00:10:19,800 --> 00:10:21,990
What's the point of this?

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00:10:21,990 --> 00:10:23,740
Well, we should maybe
have this memorized.

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00:10:23,740 --> 00:10:25,823
If you have to look it up,
you have to look it up,

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00:10:25,823 --> 00:10:27,080
but you saw this one in class.

246
00:10:27,080 --> 00:10:29,150
And the negative is
just dropping down here,

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00:10:29,150 --> 00:10:31,040
so don't think I added
that negative in when

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00:10:31,040 --> 00:10:32,360
I was taking antiderivative.

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00:10:32,360 --> 00:10:33,571
It was already there.

250
00:10:33,571 --> 00:10:34,070
All right.

251
00:10:34,070 --> 00:10:34,930
So we're done.

252
00:10:34,930 --> 00:10:36,130
Oh, but we're not done.

253
00:10:36,130 --> 00:10:37,940
Why are we not done?

254
00:10:37,940 --> 00:10:40,392
We're not done, because we
started off with something

255
00:10:40,392 --> 00:10:42,850
in terms of x, and now we have
something in terms of theta,

256
00:10:42,850 --> 00:10:44,330
so we have to finish up.

257
00:10:44,330 --> 00:10:45,770
And how we do that,
is we go back,

258
00:10:45,770 --> 00:10:47,520
we look at the
substitution we made.

259
00:10:47,520 --> 00:10:50,280
If we make a triangle
based on that substitution,

260
00:10:50,280 --> 00:10:53,320
we figure out the values of
secant theta and tangent theta,

261
00:10:53,320 --> 00:10:55,834
and then we can plug
those in terms of x.

262
00:10:55,834 --> 00:10:57,375
So let's remind
ourselves-- I'm going

263
00:10:57,375 --> 00:10:59,060
to draw the triangle
in the middle here.

264
00:10:59,060 --> 00:11:01,630
Let's remind ourselves
of the relationship

265
00:11:01,630 --> 00:11:05,100
we had between theta and x.

266
00:11:05,100 --> 00:11:10,270
If this is theta, we said
cosine theta, right here,

267
00:11:10,270 --> 00:11:12,870
cosine theta was
equal to x plus 1.

268
00:11:12,870 --> 00:11:16,300
Cosine theta is adjacent
over hypotenuse.

269
00:11:16,300 --> 00:11:20,500
So we want to say, this is
x plus 1, and this is 1.

270
00:11:20,500 --> 00:11:22,880
And that implies by the
Pythagorean theorem,

271
00:11:22,880 --> 00:11:26,180
that this is square root of
1 minus quantity x plus 1

272
00:11:26,180 --> 00:11:27,856
squared.

273
00:11:27,856 --> 00:11:29,790
Let me move that over.

274
00:11:29,790 --> 00:11:31,790
Notice, then, this
also makes sense,

275
00:11:31,790 --> 00:11:33,300
why sine theta is what it is.

276
00:11:33,300 --> 00:11:36,000
Sine theta is this
value divided by 1.

277
00:11:36,000 --> 00:11:38,601
So that also helps
you understand that.

278
00:11:38,601 --> 00:11:39,100
All right.

279
00:11:39,100 --> 00:11:40,558
So now what do we
need to read off?

280
00:11:40,558 --> 00:11:43,390
We need to read off secant, and
we need to read off tangent.

281
00:11:43,390 --> 00:11:45,790
So secant is 1 over
cosine, so actually, we

282
00:11:45,790 --> 00:11:48,640
could have gotten that one
for free, from the cosine.

283
00:11:48,640 --> 00:11:52,730
So this 1 over cosine
is 1 over x plus 1.

284
00:11:52,730 --> 00:11:56,240
So this thing is equal
to negative natural log

285
00:11:56,240 --> 00:12:01,474
absolute value 1 over x plus
1 plus-- now, what's tangent?

286
00:12:01,474 --> 00:12:03,140
If I come back and
look at the triangle,

287
00:12:03,140 --> 00:12:06,910
tangent theta is
opposite over adjacent.

288
00:12:06,910 --> 00:12:07,460
Right?

289
00:12:07,460 --> 00:12:11,380
So I can actually just put it
all over x plus 1 if I wanted.

290
00:12:11,380 --> 00:12:13,310
But I already started
writing it separately,

291
00:12:13,310 --> 00:12:15,090
so I'll leave it like this.

292
00:12:15,090 --> 00:12:20,380
Square root of 1 minus x
plus 1 quantity squared.

293
00:12:20,380 --> 00:12:23,910
And then close that,
and then my plus c.

294
00:12:23,910 --> 00:12:26,520
So now I'm actually
finished with the problem.

295
00:12:26,520 --> 00:12:31,220
Because now I have an
antiderivative in terms of x.

296
00:12:31,220 --> 00:12:33,910
So let me just remind you where
this problem, where we started

297
00:12:33,910 --> 00:12:36,100
the problem, kind of
take us through quickly,

298
00:12:36,100 --> 00:12:37,340
and then we'll be done.

299
00:12:37,340 --> 00:12:39,750
So back to the
beginning, what we had,

300
00:12:39,750 --> 00:12:45,109
was we had an integral that
was a fractional problem,

301
00:12:45,109 --> 00:12:46,650
but we had an x plus
1 here, and then

302
00:12:46,650 --> 00:12:49,850
we had this really
messy-looking quadratic in here.

303
00:12:49,850 --> 00:12:52,480
To make it easy to deal with,
I factored out a negative sign,

304
00:12:52,480 --> 00:12:54,880
and then I saw I could
complete the square.

305
00:12:54,880 --> 00:12:57,270
Once you complete the
square, you actually

306
00:12:57,270 --> 00:12:59,450
get another x plus
1 in there, which

307
00:12:59,450 --> 00:13:03,080
helps us to see immediately, it
should be a trig substitution.

308
00:13:03,080 --> 00:13:05,030
So the substitution
that's natural to make,

309
00:13:05,030 --> 00:13:07,740
because you have a 1 minus
something involving an x,

310
00:13:07,740 --> 00:13:09,980
is going to be either
cosine or sine.

311
00:13:09,980 --> 00:13:11,724
I chose cosine.

312
00:13:11,724 --> 00:13:13,140
If you'd chosen
sine, you probably

313
00:13:13,140 --> 00:13:15,270
would have gotten a
cosecant up there,

314
00:13:15,270 --> 00:13:18,804
instead of a secant, when you
were taking an antiderivative

315
00:13:18,804 --> 00:13:19,470
at the very end.

316
00:13:19,470 --> 00:13:22,280
So you would have gotten
the same answer because

317
00:13:22,280 --> 00:13:25,440
of the substitutions in the end.

318
00:13:25,440 --> 00:13:28,370
But so I chose cosine
theta is equal to x plus 1.

319
00:13:28,370 --> 00:13:29,820
You do that, you
can replace this

320
00:13:29,820 --> 00:13:33,630
with cosine, this with a sine,
this becomes a negative sine,

321
00:13:33,630 --> 00:13:35,450
and then you start simplifying.

322
00:13:35,450 --> 00:13:37,790
So once we came over
here and simplified,

323
00:13:37,790 --> 00:13:39,720
we got it into
something we recognize.

324
00:13:39,720 --> 00:13:41,140
We got it into secant.

325
00:13:41,140 --> 00:13:44,780
We know the
antiderivative for secant,

326
00:13:44,780 --> 00:13:46,110
in terms of secant and tangent.

327
00:13:46,110 --> 00:13:47,730
We know it's exactly this.

328
00:13:47,730 --> 00:13:50,730
And then we went back to
the relationship we had.

329
00:13:50,730 --> 00:13:53,470
We made ourselves
a triangle in terms

330
00:13:53,470 --> 00:13:55,130
of the theta and the x-values.

331
00:13:55,130 --> 00:13:57,130
And then we were
able to substitute in

332
00:13:57,130 --> 00:13:59,300
for secant and tangent.

333
00:13:59,300 --> 00:14:00,040
All right.

334
00:14:00,040 --> 00:14:02,140
So hopefully that was
successful for you.

335
00:14:02,140 --> 00:14:04,145
And that's where I'll stop.