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Hi.

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Welcome back to recitation.

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We've been talking about a
bunch of different integration

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techniques.

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Here are a couple of examples on
which you can try and pick out

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the right technique to compute
these two integrals with.

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So I've got the first one,
is the indefinite integral

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of x cubed times the
square root of the quantity

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x squared plus 2 dx.

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And the second one is
a definite integral.

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It's the integral
1 to e of x squared

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times the quantity ln of 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 compute these two
integrals, come back,

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and we can compute
them together.

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So hopefully you had some luck
working on these integrals.

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Let's start with the first one.

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So for question A, let
me just rewrite it here.

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We have the integral of x cubed
times the square root of x

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squared plus 2 dx.

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Now, if you look at this
integral, what you see

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is a product of two terms.

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This term is x cubed.

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It's a nice term.

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This one is the square
root of x squared plus 2.

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That's not such a nice term.

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There are a few different things
you might think to try here.

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So one thing you
might think to try,

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is this square root
of x squared plus 2 is

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the sort of thing that shows
up in trig substitutions.

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So that's one thing you
might have in your mind.

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Another thing you might
have in your mind,

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is that this is a
product of two things.

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So you might consider
the possibility

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that you're going to do an
integration by parts here.

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And the third thing you might
have in your mind is just,

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you could just-- you
know, this might just

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be a substitution situation.

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You have this x squared plus 2
as the sort of discrete entity

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that shows up there.

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So you might try a
regular substitution.

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Now you have to, you know, make
a choice among which of these

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to try.

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And it seems to me
that of those options,

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a regular substitution
is the easiest to do.

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And you'll find
out really quickly

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whether you've made your life
horribly complicated or not.

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And frequently, with
integration by parts,

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if you have the option of doing
a substitution when you have

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integration by parts, you
know, it won't-- if you do

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the substitution, you still
can do integration by parts

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afterward.

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If you do integration
by parts first,

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you can still do the
substitution afterward.

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So trying the substitution
shouldn't hurt us

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for integration by parts,
and a regular substitution

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will be a lot simpler than a
trig substitution here, right?

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Trig substitutions in
general are complicated,

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so it's a good idea to see if
there's something simpler that

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can be done first.

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So if we're going to try
regular substitution,

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the sort of natural
thing to do, is

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to set u equal x squared plus 2.

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So in this case, we have
du is equal to 2x dx.

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Or-- well, so OK.

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So we want to substitute
these things in.

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The u is going to go into
the square root term.

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So that's going to
be taken care of.

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And the dx part--
well, we've got an x

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here to pair up with the dx, so
that's going to give us a du.

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And then we'll have
x squared left over.

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So we'll need to substitute
in for x squared.

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So we're going to use x
squared is equal to u minus 2,

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and x dx is equal to du over 2.

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So when we make these
substitutions in here, what

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we get is that this integral
is equal to the integral

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of-- so it's x squared times
the square root of x squared

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plus 2 times x dx.

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So that's u minus 2
times the square root

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of u times du over 2.

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Now, having made the
substitution, we see-- OK.

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So with a little bit
of rearrangement,

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this is just a sum
of powers of u.

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Right?

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We can rewrite this as the
integral of u to the 3/2 over 2

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minus u to the 1/2 du.

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And now, this is just
straightforward to integrate.

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So it turns out we don't
need integration by parts.

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We don't need a
trig substitution.

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Just this regular
substitution works nicely.

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So we get, well, u
to the 3/2, so we

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compute an antiderivative, so
that gives us u to the 5/2.

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And I have to divide by 5/2,
so I have to multiply by 2/5.

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And then I have a half.

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So it's u to the 5/2 over
5, minus-- so u to the 1/2,

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so that gives me 3/2, so I
need minus 2/3 u to the 3/2.

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OK.

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And now to finish
this off, I have

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to do my final
back substitution,

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and replace u with x.

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So this is going to be
x squared plus 2 over 5

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to the 5/2 minus 2/3 times
x squared plus 2 to the 3/2.

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All right.

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Plus a constant.

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Thank you, peanut gallery.

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Plus a constant.

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Right.

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Indefinite integral,
should have had it

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right the very first
time I went through.

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Plus a constant, both times.

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OK.

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Good.

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So that's all there
is for part A. Now

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let's look at part B. So OK.

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So part B is a little
more complicated-looking.

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Let me copy it over here again.

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So part B, we have the
integral from 1 to e

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of x squared times the
quantity ln x squared dx.

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So again, there are a couple
of possibilities for what

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we could want to do here.

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One thing is you could try
just a regular substitution.

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So maybe a regular
substitution you

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might want to try is u equals
ln of x, or x equals e to the u.

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And now if you try that, you
can try it, and what you'll see,

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is you'll get something.

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So here we have a
polynomial times a function

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of the logarithm, a
power of the logarithm.

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If you try that substitution,
what you'll end up with

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is a polynomial in u
times an exponential in u.

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So OK.

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So it's not clear that
we've won anything,

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if we make that substitution.

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The other thing we can try,
is we can try integration

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by parts here.

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Again, we have a product form.

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And if you make
that substitution,

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you get polynomial
times exponential,

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and that's also a promising
place for integration by parts,

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after the substitution.

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So OK.

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So I'm not going to
make the substitution.

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I'm just going to go--
seeing this product,

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I'm going to go for
integration by parts directly.

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So I need to decide what pieces
are which for integration

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by parts.

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And the thing to
remember is this

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that-- so I have here
a polynomial times

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a function of the logarithm.

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And the thing to remember
is that polynomials

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like to be differentiated.

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That's their preference.

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But logarithms really, really
like to be differentiated.

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You'd much rather differentiate
a logarithm than integrate it.

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You can integrate a
polynomial-- I mean you

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can integrate either of them.

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You can integrate a polynomial.

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When you integrate
a polynomial, you

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get something that's still
a fairly nice polynomial.

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Integrating
logarithms is harder,

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whereas differentiating
logarithms makes them

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into just powers of x.

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So setting this up as
an integration by parts,

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we're going to
want to take this,

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the everything with
the logarithm in it,

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and that's the piece we
want to differentiate,

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to try and simplify it.

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So that's going
to be our u part.

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So we're going to try
u equals ln x squared,

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and that leaves v
prime to be x squared.

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So to do integration
by parts-- OK.

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We need to differentiate
and we need to integrate.

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So if I differentiate ln
x squared, I get u prime.

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The prime is equal to-- OK.

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Well, this is a
little chain rule.

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So it's 2 ln x times
the derivative of ln

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x, which is 1 over x.

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And if I integrate
v prime, I get

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v is equal to x cubed over 3.

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OK.

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So now I use the formula
for integration by parts.

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So I have that this integral--
what I'm going to do is

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I'm going to call this,
let me give it a name, I.

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Now I'm going to come down
here and I'm going to say,

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I is equal to-- well,
by integration by parts,

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it's equal to u*v minus the
integral of u prime v dx.

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So in our case, u*v is x cubed
over 3 times ln x squared.

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But the thing to remember is,
this was a definite integral

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to start out with.

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It's from 1 to e.

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So when you take
this part out front,

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you have to take the
difference of its two values.

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So we're going to evaluate
this between x equals 1

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and x equals e, minus
the definite integral--

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so we still have a definite
integral from one to e

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of u prime v. So
that's 2/3-- OK.

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So now we have an ln x
over x times x cubed.

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So the x and the x cubed,
that cancels one power

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of x out of the numerator.

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So we're left with
2/3 x squared ln x dx.

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So.

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One thing we see,
this is just going

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to be-- this is just going
to evaluate to a constant.

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We're going to plug in e,
and we're going to plug in 1,

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and we're going to
take the difference.

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So that's just some number.

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This part is still
some integral.

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Are we done yet?

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Well, it's not obvious
immediately how

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to integrate this.

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I mean, I don't already
know off the top of my head

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what the antiderivative here is.

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On the other hand,
it's definitely

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simpler than what we
started with, right?

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Before we had x
squared ln x squared.

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Here we just have
x squared ln x.

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So we've simplified
the integrand.

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And it's still in a form in
which we can further integrate

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by parts, if you wanted to.

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Right?

220
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So now in this case
the natural thing to do

221
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would be very similar
to what we did before.

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We're going to take
the u to be the ln

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x, to be the piece
we differentiate,

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and we're going to take
the v to be x squared.

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OK?

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So we can do that.

227
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So this is equal to-- let's see.

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I'm going to bring
it, now, upstairs.

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So this is equal to-- all right.

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So we plug in e here.

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We get e cubed over 3 times
1, minus when we plug in 1,

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ln of 1 is 0.

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So that just gives
me e cubed over 3

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for the first part
minus-- all right,

235
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and I'm going to pull
the 2/3 out front.

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So it's minus 2/3 times--

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OK.

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So now I'm going to
do an integration

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by parts on this one.

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So again, like I said, I'm going
to take u is equal to ln x.

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I'm going to take v prime
is equal to x squared.

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So that means u
prime is equal to 1

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00:11:05,510 --> 00:11:13,190
over x and v is equal
to x cubed over 3.

244
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And so again, I get uv,
which is x cubed over 3 ln x,

245
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and again, because it
was a definite integral,

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I have to evaluate this between
x equals 1 and e, minus-- OK.

247
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So minus the integral
of u prime v.

248
00:11:31,580 --> 00:11:34,050
So this is the
integral from 1 to e.

249
00:11:34,050 --> 00:11:36,560
u prime v is 1 over x
times x cubed over 3,

250
00:11:36,560 --> 00:11:43,070
so that's x squared over 3 dx.

251
00:11:43,070 --> 00:11:45,345
OK, so having written all
that down, what you see

252
00:11:45,345 --> 00:11:47,440
is this is a constant minus 2/3.

253
00:11:47,440 --> 00:11:47,940
OK.

254
00:11:47,940 --> 00:11:48,860
This is just some constant.

255
00:11:48,860 --> 00:11:50,250
We'll plug in e,
we'll plug in 1.

256
00:11:50,250 --> 00:11:50,920
That's easy.

257
00:11:50,920 --> 00:11:53,500
And now we have an integral,
and we've reduced this finally.

258
00:11:53,500 --> 00:11:55,699
Here we have just an
integral of a polynomial.

259
00:11:55,699 --> 00:11:57,990
So an integral of a polynomial,
that's easy to compute,

260
00:11:57,990 --> 00:11:59,900
you can do it yourself.

261
00:11:59,900 --> 00:12:03,830
And I'm going to let you finish
off the last couple of steps

262
00:12:03,830 --> 00:12:04,820
for yourself.

263
00:12:04,820 --> 00:12:07,540
In the end, the
answer should work out

264
00:12:07,540 --> 00:12:19,900
to 5 e cubed over
27 minus 2 over 27.

265
00:12:19,900 --> 00:12:22,140
This is what you should
find the final answer

266
00:12:22,140 --> 00:12:23,580
to be when you work it all out.

267
00:12:23,580 --> 00:12:25,420
So I'll end there.