WEBVTT

00:00:06.920 --> 00:00:07.430
Hi.

00:00:07.430 --> 00:00:08.874
Welcome back to recitation.

00:00:08.874 --> 00:00:10.540
In lecture, you've
learned a whole bunch

00:00:10.540 --> 00:00:12.290
of different
integration techniques.

00:00:12.290 --> 00:00:13.970
So I have here a
couple of problems

00:00:13.970 --> 00:00:15.490
that will give
you an opportunity

00:00:15.490 --> 00:00:19.080
to apply those techniques, and
figure out which ones to apply.

00:00:19.080 --> 00:00:22.400
So in particular I have
one definite integral,

00:00:22.400 --> 00:00:26.085
the integral 0 to 2 of
the quantity x times e

00:00:26.085 --> 00:00:28.710
to the 1 minus x squared
with respect to x.

00:00:28.710 --> 00:00:31.150
And the second one is
an indefinite integral

00:00:31.150 --> 00:00:35.800
of 2 arctan x divided by x
squared with respect to x.

00:00:35.800 --> 00:00:38.321
So why don't you pause
the video, take some time

00:00:38.321 --> 00:00:40.570
to work these out, come back,
and we can work them out

00:00:40.570 --> 00:00:41.069
together.

00:00:49.630 --> 00:00:50.400
Welcome back.

00:00:50.400 --> 00:00:52.710
Hopefully you had some luck
working on these integrals.

00:00:52.710 --> 00:00:53.460
Let's get started.

00:00:53.460 --> 00:00:55.335
We can do the first one
first, and then we'll

00:00:55.335 --> 00:00:56.750
go on to the second one.

00:00:56.750 --> 00:01:02.573
So here we want to compute the
integral from 0 to 2 of x e

00:01:02.573 --> 00:01:06.700
to the 1 minus x squared dx.

00:01:06.700 --> 00:01:10.020
So in order to do this, we
have to, you know-- well,

00:01:10.020 --> 00:01:12.682
we could look at it and say, do
I know the answer immediately,

00:01:12.682 --> 00:01:13.640
off the top of my head?

00:01:13.640 --> 00:01:16.240
And this is not one
of the, you know,

00:01:16.240 --> 00:01:19.215
ones that we have
immediately already memorized

00:01:19.215 --> 00:01:20.430
a formula for.

00:01:20.430 --> 00:01:21.350
So then you say, OK.

00:01:21.350 --> 00:01:24.069
So now I need to do something
in order to try to integrate it.

00:01:24.069 --> 00:01:26.360
And so, you know, you have
a whole bunch of techniques.

00:01:26.360 --> 00:01:28.529
And for this one,
when you look at it,

00:01:28.529 --> 00:01:30.320
the technique that is
going to work for you

00:01:30.320 --> 00:01:31.880
is just the simplest
technique that you have,

00:01:31.880 --> 00:01:33.810
which is just a
straightforward substitution.

00:01:33.810 --> 00:01:36.960
And one way to figure that out,
is that you can look at this

00:01:36.960 --> 00:01:38.530
and you can see
that here we have

00:01:38.530 --> 00:01:41.286
a composition of functions,
this e to the 1 minus x

00:01:41.286 --> 00:01:42.660
squared, and so
whenever you have

00:01:42.660 --> 00:01:44.118
a composition of
functions, there's

00:01:44.118 --> 00:01:46.430
the opportunity for there
to have been a chain

00:01:46.430 --> 00:01:47.610
rule somewhere.

00:01:47.610 --> 00:01:50.630
And in this case, we see that
this composition of functions

00:01:50.630 --> 00:01:55.290
is multiplied by x, and
x is very closely related

00:01:55.290 --> 00:01:58.890
to the derivative of
1 minus x squared.

00:01:58.890 --> 00:02:02.720
So that's what's suggesting
for us a nice substitution.

00:02:02.720 --> 00:02:04.810
And so the substitution
we're going to try, then,

00:02:04.810 --> 00:02:06.942
is that we're going to
put this inside function,

00:02:06.942 --> 00:02:09.400
this 1 minus x squared, we're
going to set u equal to that.

00:02:09.400 --> 00:02:09.983
So, all right.

00:02:09.983 --> 00:02:12.960
So this is equal to--
so I'm going to say,

00:02:12.960 --> 00:02:15.870
u equal 1 minus x squared.

00:02:15.870 --> 00:02:19.230
So now if I take a differential,
if u equals 1 minus x squared,

00:02:19.230 --> 00:02:25.677
that means that
du is minus 2x dx.

00:02:25.677 --> 00:02:27.260
And since this is a
definite integral,

00:02:27.260 --> 00:02:29.630
I also need to change the
bounds of integration.

00:02:29.630 --> 00:02:34.620
So when x is equal to 0, that
means that u is equal to,

00:02:34.620 --> 00:02:38.740
well, 1 minus 0 squared is
1, so that's the lower bound.

00:02:38.740 --> 00:02:41.240
And when x is equal
to 2, that means

00:02:41.240 --> 00:02:45.600
that u is equal to 1 minus 2
squared, which is negative 3.

00:02:45.600 --> 00:02:46.892
So we make these substitutions.

00:02:46.892 --> 00:02:47.391
OK.

00:02:47.391 --> 00:02:48.950
So what does our
integral become?

00:02:48.950 --> 00:02:52.950
So this becomes the integral--
so now the lower bound

00:02:52.950 --> 00:02:55.985
becomes u equals 1,
and the upper bound

00:02:55.985 --> 00:02:58.450
is u equals negative 3.

00:02:58.450 --> 00:02:58.950
OK.

00:02:58.950 --> 00:03:01.260
So now e to the 1
minus x squared,

00:03:01.260 --> 00:03:02.580
that's just e to the u.

00:03:05.480 --> 00:03:08.940
And x dx-- well,
that's almost du.

00:03:08.940 --> 00:03:11.840
It's du divided by negative 2.

00:03:11.840 --> 00:03:17.750
So du divided by negative 2.

00:03:17.750 --> 00:03:19.510
So we make the substitution.

00:03:19.510 --> 00:03:21.259
We transform our
integral to this form.

00:03:21.259 --> 00:03:23.300
Now, this is a very nice,
simple integral, right?

00:03:23.300 --> 00:03:25.780
This is just the integral of
e to the u with a, you know,

00:03:25.780 --> 00:03:26.772
a constant multiple.

00:03:26.772 --> 00:03:27.272
So, OK.

00:03:27.272 --> 00:03:29.030
So we can do this right away.

00:03:29.030 --> 00:03:32.900
So this is-- I'll pull the
minus 1/2 out in front.

00:03:32.900 --> 00:03:35.340
And now fundamental
theorem of calculus.

00:03:35.340 --> 00:03:38.620
Antiderivative of the e to
the u is just e to the u.

00:03:38.620 --> 00:03:40.620
And I have to take
that between my bounds.

00:03:40.620 --> 00:03:45.542
So between u equals 1
and u equals minus 3.

00:03:45.542 --> 00:03:47.250
So notice that because
we changed bounds,

00:03:47.250 --> 00:03:50.770
we don't have to go back and
convert everything back to x.

00:03:50.770 --> 00:03:53.340
We can just plug in directly.

00:03:53.340 --> 00:03:53.840
So OK.

00:03:53.840 --> 00:03:55.060
So we plug in directly.

00:03:55.060 --> 00:03:55.690
What do I get?

00:03:55.690 --> 00:03:59.447
So first I get minus
1/2 times-- all right,

00:03:59.447 --> 00:04:01.280
so I'll just pull the
constant out in front.

00:04:01.280 --> 00:04:09.430
So the first term is e to the
minus third minus e to the 1.

00:04:09.430 --> 00:04:09.930
OK.

00:04:09.930 --> 00:04:13.660
And if I wanted to, I could
rewrite this as e minus e

00:04:13.660 --> 00:04:18.430
to the minus 3 over 2, or
in a variety of other ways.

00:04:18.430 --> 00:04:18.930
All right.

00:04:18.930 --> 00:04:22.300
So this one was a pretty
straightforward integration

00:04:22.300 --> 00:04:23.470
by substitution.

00:04:23.470 --> 00:04:25.920
Definite integral, change
bounds, and at the end

00:04:25.920 --> 00:04:27.770
you get this answer.

00:04:27.770 --> 00:04:29.480
So let's take a look
at the second one.

00:04:29.480 --> 00:04:31.460
So the second one is a
bit trickier, I think,

00:04:31.460 --> 00:04:32.970
and it requires a bit more work.

00:04:32.970 --> 00:04:34.111
So here it is.

00:04:34.111 --> 00:04:34.860
Let me rewrite it.

00:04:34.860 --> 00:04:45.910
So it's the antiderivative of
2 arctan x over x squared dx.

00:04:45.910 --> 00:04:49.170
So we have here
an antiderivative

00:04:49.170 --> 00:04:50.440
that we have to compute.

00:04:50.440 --> 00:04:52.786
And this is a kind of
messy-looking expression

00:04:52.786 --> 00:04:53.785
that we're working with.

00:04:53.785 --> 00:04:55.140
All right?

00:04:55.140 --> 00:04:59.230
So this is not very pleasant.

00:04:59.230 --> 00:05:01.560
So there are a couple
of different things

00:05:01.560 --> 00:05:02.660
we could think to do.

00:05:02.660 --> 00:05:07.490
So one thing that you could
think to do is you could say,

00:05:07.490 --> 00:05:09.580
I really don't like arctan.

00:05:09.580 --> 00:05:11.431
I'm going to try and
get rid of arctan.

00:05:11.431 --> 00:05:13.180
And the way I'm going
to get rid of arctan

00:05:13.180 --> 00:05:15.950
is by doing a substitution,
u equals arctan x,

00:05:15.950 --> 00:05:17.900
or x equals tan u.

00:05:17.900 --> 00:05:19.320
So if you do that,
that'll get rid

00:05:19.320 --> 00:05:22.110
of your inverse trigonometric
function, but it'll introduce,

00:05:22.110 --> 00:05:23.920
in the bottom, you
have this x squared,

00:05:23.920 --> 00:05:27.180
so that'll introduce
like a tan u squared,

00:05:27.180 --> 00:05:30.057
and dx will give you
some more trig functions.

00:05:30.057 --> 00:05:31.640
So what you'll be
left with, then-- so

00:05:31.640 --> 00:05:33.240
this will be u on top.

00:05:33.240 --> 00:05:36.559
So you'll have u times a
product of trig functions.

00:05:36.559 --> 00:05:38.100
So I guess it's up
to you whether you

00:05:38.100 --> 00:05:41.240
think that a polynomial,
or a power of u

00:05:41.240 --> 00:05:44.720
times a bunch of trig functions,
is simpler than an inverse trig

00:05:44.720 --> 00:05:47.510
function times a power.

00:05:47.510 --> 00:05:51.431
It's not clear to me that it's
actually that much simpler.

00:05:51.431 --> 00:05:52.930
And in any case,
we wouldn't sort of

00:05:52.930 --> 00:05:54.030
be finished at that stage.

00:05:54.030 --> 00:05:56.560
There would still be a fair
amount more work to do.

00:05:56.560 --> 00:05:59.170
So we can think about, what
other things could we do?

00:05:59.170 --> 00:06:01.860
That's kind of the only
obvious substitution here.

00:06:01.860 --> 00:06:05.750
But another thing
that suggests itself,

00:06:05.750 --> 00:06:08.450
is that we could try an
integration by parts.

00:06:08.450 --> 00:06:10.980
And one reason is, this
is a product of things.

00:06:10.980 --> 00:06:13.840
You know, here it's arctan
x times 1 over x squared.

00:06:13.840 --> 00:06:18.170
And arctan x is something that
works nicely with integration

00:06:18.170 --> 00:06:19.880
by parts if you can
differentiate it.

00:06:19.880 --> 00:06:20.380
Right?

00:06:20.380 --> 00:06:22.550
So inverse
trigonometric functions

00:06:22.550 --> 00:06:24.810
behave like logarithms
with respect

00:06:24.810 --> 00:06:26.430
to integration by parts.

00:06:26.430 --> 00:06:28.832
By which I mean, they really
like to be differentiated.

00:06:28.832 --> 00:06:30.540
Because when you
differentiate an arctan,

00:06:30.540 --> 00:06:32.020
you get something much simpler.

00:06:32.020 --> 00:06:34.240
Well, simpler, anyhow.

00:06:34.240 --> 00:06:35.680
Maybe not much simpler.

00:06:35.680 --> 00:06:37.670
You get 1 over 1 plus x squared.

00:06:37.670 --> 00:06:39.390
So if you can apply
integration by parts

00:06:39.390 --> 00:06:42.590
and differentiate the arctan,
that makes your life nice.

00:06:42.590 --> 00:06:44.520
And so, then, in order
for that to work,

00:06:44.520 --> 00:06:46.485
you need for everything
else to be something

00:06:46.485 --> 00:06:49.430
you can antidifferentiate, and
here everything else is just 2

00:06:49.430 --> 00:06:51.200
over x squared, and
so that's something

00:06:51.200 --> 00:06:52.590
that we know how to integrate.

00:06:52.590 --> 00:06:53.430
It's easy to do.

00:06:53.430 --> 00:06:53.930
So, OK.

00:06:53.930 --> 00:06:56.289
So let's give that a try, then.

00:06:56.289 --> 00:06:58.580
So we're going to do integration
by parts, so I'm going

00:06:58.580 --> 00:07:06.000
to set u equal to arctan x.

00:07:06.000 --> 00:07:12.510
So that means u prime is equal
to 1 over 1 plus x squared.

00:07:12.510 --> 00:07:15.001
And I'm going to
let v prime equal,

00:07:15.001 --> 00:07:17.000
well, I guess I have to
pick up the 2 somewhere,

00:07:17.000 --> 00:07:18.590
so I might as well
pick it up here.

00:07:18.590 --> 00:07:22.090
2 over x squared.

00:07:22.090 --> 00:07:25.680
And so then v, so I
integrate 2 over x squared,

00:07:25.680 --> 00:07:30.530
so that gives me x to the
minus 1, with a minus sign.

00:07:30.530 --> 00:07:34.740
So it's minus 2 over x.

00:07:34.740 --> 00:07:37.230
So these are u, u
prime, v and v prime.

00:07:37.230 --> 00:07:40.700
So now OK, so now I apply
integration by parts,

00:07:40.700 --> 00:07:44.780
and that tells me this is
equal to-- well, let's see.

00:07:44.780 --> 00:07:54.425
So it's u*v, which is
minus 2 arctan x over x.

00:07:54.425 --> 00:08:00.570
That's u*v. Minus the integral
of, now it's the second part,

00:08:00.570 --> 00:08:04.600
it's v u prime dx.

00:08:04.600 --> 00:08:07.560
So here v u prime is-- well, OK.

00:08:07.560 --> 00:08:09.190
So they're minus 2.

00:08:09.190 --> 00:08:11.300
I'm going to pull
that out front.

00:08:11.300 --> 00:08:13.450
So it's plus 2-- all right.

00:08:13.450 --> 00:08:18.250
And then the rest of
it is 1 over x times 1

00:08:18.250 --> 00:08:21.940
plus x squared dx.

00:08:21.940 --> 00:08:23.930
So if you apply
integration by parts,

00:08:23.930 --> 00:08:26.489
you have the u*v stuff that
gets kicked out out front,

00:08:26.489 --> 00:08:28.780
and then what you're left
with is this second integral.

00:08:28.780 --> 00:08:32.394
Now this is, to me this is
simpler-looking than what

00:08:32.394 --> 00:08:33.060
we started with.

00:08:33.060 --> 00:08:35.520
It's still not simple,
but it's simpler.

00:08:35.520 --> 00:08:38.296
So it seems to me that,
you know, having reached

00:08:38.296 --> 00:08:39.420
this stage, we can say, OK.

00:08:39.420 --> 00:08:42.391
We've made some progress, and
now we have to keep going.

00:08:42.391 --> 00:08:42.890
Right?

00:08:42.890 --> 00:08:45.720
So it's not obvious to me
what the antiderivative

00:08:45.720 --> 00:08:48.340
of this expression should be.

00:08:48.340 --> 00:08:49.640
So how can I figure that out?

00:08:49.640 --> 00:08:52.230
Well, there are a couple
of different options here.

00:08:52.230 --> 00:08:54.900
One thing you might
look at this and see,

00:08:54.900 --> 00:08:56.950
is you might see--
again, I have a 1

00:08:56.950 --> 00:08:58.720
plus x squared in
the denominator,

00:08:58.720 --> 00:09:00.620
and so that might
make you tempted to do

00:09:00.620 --> 00:09:02.950
a trig substitution again.

00:09:02.950 --> 00:09:05.390
And in fact, you could do
a trig substitution here.

00:09:05.390 --> 00:09:08.360
You could put x equal
tan theta, and you

00:09:08.360 --> 00:09:10.754
would be able to solve
the question like this.

00:09:10.754 --> 00:09:12.670
But I don't think it's
the simplest way to go.

00:09:12.670 --> 00:09:14.787
Because in addition--
well, on the one hand,

00:09:14.787 --> 00:09:16.870
this has this 1 plus x
squared in the denominator,

00:09:16.870 --> 00:09:19.400
but on the other hand, this
is just a rational function.

00:09:19.400 --> 00:09:19.900
Yeah?

00:09:19.900 --> 00:09:21.580
It's a ratio of two polynomials.

00:09:21.580 --> 00:09:23.720
The polynomial on top is
just 1, and the polynomial

00:09:23.720 --> 00:09:26.950
on the bottom is this product,
x times 1 plus x squared.

00:09:26.950 --> 00:09:29.450
And so whenever you have a
rational function that you're

00:09:29.450 --> 00:09:30.830
trying to integrate,
you can also

00:09:30.830 --> 00:09:35.950
always use-- what's it called--
partial fraction decomposition.

00:09:35.950 --> 00:09:37.590
So we can try to use
partial fractions

00:09:37.590 --> 00:09:39.290
on the second expression.

00:09:39.290 --> 00:09:43.140
And so I think I'd
like to do it that way.

00:09:43.140 --> 00:09:44.760
So let's do that.

00:09:44.760 --> 00:09:45.760
Let's do it down here.

00:09:45.760 --> 00:09:56.420
So partial fractions
on the quantity 1

00:09:56.420 --> 00:10:01.722
over x times 1 plus x squared.

00:10:01.722 --> 00:10:03.680
So if you remember your
partial fractions here,

00:10:03.680 --> 00:10:05.090
this is completely factored.

00:10:05.090 --> 00:10:07.080
This quadratic doesn't factor.

00:10:07.080 --> 00:10:10.150
It doesn't have any real
roots, doesn't factor.

00:10:10.150 --> 00:10:13.960
So OK, so we have a
single linear term,

00:10:13.960 --> 00:10:16.340
and a single non-factorable
quadratic term.

00:10:16.340 --> 00:10:18.940
So partial fractions tells
us that we can write this

00:10:18.940 --> 00:10:24.880
in the form A over x plus-- and
now because the second term is

00:10:24.880 --> 00:10:31.210
quadratic, it needs to be B*x
plus C over 1 plus x squared.

00:10:31.210 --> 00:10:31.710
Right?

00:10:31.710 --> 00:10:34.120
So when you have a
quadratic in the bottom,

00:10:34.120 --> 00:10:38.510
you get two constants up at
the top, a linear polynomial.

00:10:38.510 --> 00:10:39.760
And then you can always check.

00:10:39.760 --> 00:10:43.130
You have three constants
here, and the degree

00:10:43.130 --> 00:10:44.690
of the denominator
is 3, so that's

00:10:44.690 --> 00:10:47.700
one way of checking that
you're not completely off base,

00:10:47.700 --> 00:10:49.400
that the degree down
here should always

00:10:49.400 --> 00:10:50.816
agree with the
number of constants

00:10:50.816 --> 00:10:52.390
that you're solving for.

00:10:52.390 --> 00:10:53.380
OK.

00:10:53.380 --> 00:10:53.880
Good.

00:10:53.880 --> 00:10:57.310
So now we need to solve
this for A, B, and C.

00:10:57.310 --> 00:11:00.620
And so OK, because we have
this nice single linear factor,

00:11:00.620 --> 00:11:02.730
we can do the cover
up method there.

00:11:02.730 --> 00:11:07.776
So we cover up x, and
we cover up over x,

00:11:07.776 --> 00:11:09.384
and we cover up everything else.

00:11:09.384 --> 00:11:10.800
So on the right-hand
side, we just

00:11:10.800 --> 00:11:14.290
end up with A, and on the
left-hand side, well, so x.

00:11:14.290 --> 00:11:16.850
So that's what we need, we
get whatever when we plug in

00:11:16.850 --> 00:11:17.960
x equals 0.

00:11:17.960 --> 00:11:21.900
So on this side, we get
1 over 1 plus 0 squared,

00:11:21.900 --> 00:11:25.060
so that's just 1.

00:11:25.060 --> 00:11:29.770
So the cover up method gives
us that A is equal to 1.

00:11:29.770 --> 00:11:31.630
And now we need to
solve for B and C.

00:11:31.630 --> 00:11:33.590
And probably the most
straightforward way,

00:11:33.590 --> 00:11:35.190
in this case, to
solve for B and C,

00:11:35.190 --> 00:11:39.340
is you can always
just multiply through.

00:11:39.340 --> 00:11:40.997
And if you multiply
through-- OK.

00:11:40.997 --> 00:11:42.830
So we'll multiply through
on the left-hand--

00:11:42.830 --> 00:11:44.830
we'll clear denominators,
we'll multiply through

00:11:44.830 --> 00:11:46.500
by x times 1 plus x squared.

00:11:46.500 --> 00:11:49.240
So on the left we'll get just 1.

00:11:49.240 --> 00:11:50.350
On the right-- OK.

00:11:50.350 --> 00:11:56.830
So we multiply A over x times
x times 1 plus x squared,

00:11:56.830 --> 00:11:58.890
so that gives us A
times 1 plus x squared,

00:11:58.890 --> 00:12:00.990
but we know that A
is 1, so this is--

00:12:00.990 --> 00:12:05.080
the first term becomes
1 plus x squared,

00:12:05.080 --> 00:12:08.560
and the second term we multiply
through by x times 1 plus x

00:12:08.560 --> 00:12:10.700
squared, and the 1
plus x squareds cancel,

00:12:10.700 --> 00:12:18.850
and we're left with
plus x times B*x plus C.

00:12:18.850 --> 00:12:22.820
Now, if you, maybe you
could-- at this point,

00:12:22.820 --> 00:12:24.810
there are a number of
different ways to finish.

00:12:24.810 --> 00:12:27.040
But one, for example, is
you could see, all right,

00:12:27.040 --> 00:12:30.970
if you subtract the 1 plus x
squared over to the other side,

00:12:30.970 --> 00:12:35.230
you have minus x
squared, and over here,

00:12:35.230 --> 00:12:38.540
you have B x squared plus C*x.

00:12:38.540 --> 00:12:43.040
So for those things to be equal,
you need B to be negative 1,

00:12:43.040 --> 00:12:45.910
and you need C to be 0.

00:12:45.910 --> 00:12:47.256
Right?

00:12:47.256 --> 00:12:47.756
So OK.

00:12:47.756 --> 00:12:51.700
So good.

00:12:51.700 --> 00:12:57.817
So we have B equals
minus 1, C equals 0.

00:12:57.817 --> 00:12:59.400
So the partial
fraction decomposition,

00:12:59.400 --> 00:13:07.830
then, if we plug this in,
is 1 over x minus x over 1

00:13:07.830 --> 00:13:12.300
plus x squared.

00:13:12.300 --> 00:13:14.720
So this is the partial
fraction decomposition

00:13:14.720 --> 00:13:18.890
of this rational function.

00:13:18.890 --> 00:13:20.980
So you apply that partial
fraction decomposition.

00:13:20.980 --> 00:13:21.605
What do you do?

00:13:21.605 --> 00:13:24.640
All right, So now you carry that
back upstairs to the integral

00:13:24.640 --> 00:13:25.660
we were working on.

00:13:29.170 --> 00:13:31.700
So our integral
that we started with

00:13:31.700 --> 00:13:34.910
is equal to, now-- well, so this
constant is still out in front.

00:13:34.910 --> 00:13:45.420
So it's minus 2 arctan
x over x plus-- so

00:13:45.420 --> 00:13:47.370
now we've got 2
times-- all right.

00:13:47.370 --> 00:13:49.700
So first, this is just
algebra that we've done,

00:13:49.700 --> 00:13:51.210
so we can just substitute it in.

00:13:51.210 --> 00:13:54.200
So we replace this
1 over x times 1

00:13:54.200 --> 00:14:04.910
plus x squared by 1 over x minus
x over 1 plus x squared dx.

00:14:04.910 --> 00:14:05.410
OK.

00:14:05.410 --> 00:14:07.810
And now using the
properties of integration,

00:14:07.810 --> 00:14:10.510
this is a difference of two--
an integral of a difference

00:14:10.510 --> 00:14:13.440
is the difference of
the two integrals.

00:14:13.440 --> 00:14:15.780
So we just integrate
them separately, right?

00:14:15.780 --> 00:14:19.460
So this is equal to-- so OK, the
part out front never changes.

00:14:19.460 --> 00:14:26.580
2 arctan x over x,
plus 2 times-- OK.

00:14:26.580 --> 00:14:30.200
So you integrate 1 over
x, and that just gives you

00:14:30.200 --> 00:14:34.180
ln of absolute value of x.

00:14:34.180 --> 00:14:38.280
And OK, so you integrate-- so
the second part is x over 1

00:14:38.280 --> 00:14:39.300
plus x squared.

00:14:39.300 --> 00:14:41.750
Now, in the worst
case, we would need

00:14:41.750 --> 00:14:44.062
to do some more work on this
term and split things up.

00:14:44.062 --> 00:14:46.270
And one of them, we might
have to complete the square

00:14:46.270 --> 00:14:46.820
or something.

00:14:46.820 --> 00:14:49.320
But in this case, it actually
has worked out kind of nicely.

00:14:49.320 --> 00:14:54.850
This x over 1 plus x squared
is a simple thing to handle.

00:14:54.850 --> 00:14:56.860
That's just ln of
1 plus x squared.

00:14:56.860 --> 00:15:00.410
Well, actually, one half
of ln of 1 plus x squared.

00:15:00.410 --> 00:15:09.500
So this is minus 1/2
ln of 1 plus x squared.

00:15:09.500 --> 00:15:11.640
I could write absolute
value, but 1 plus x squared

00:15:11.640 --> 00:15:15.517
is always positive, so it
doesn't matter if I do or not.

00:15:15.517 --> 00:15:17.350
And if I wanted, I could
rewrite this a bit.

00:15:17.350 --> 00:15:20.356
I could rewrite this
as-- just to finish it,

00:15:20.356 --> 00:15:27.200
I could write 2
arctan x over x plus 2

00:15:27.200 --> 00:15:36.550
ln absolute value of x minus
ln of 1 plus x squared.

00:15:36.550 --> 00:15:38.746
And I was doing
an antiderivative.

00:15:38.746 --> 00:15:40.870
So at some point, whenever
I finished the last one,

00:15:40.870 --> 00:15:42.161
I should have added a constant.

00:15:42.161 --> 00:15:44.440
So this should have
had a plus c here,

00:15:44.440 --> 00:15:48.050
and it should have
a plus c there.

00:15:48.050 --> 00:15:51.140
Right, good.

00:15:51.140 --> 00:15:54.350
So OK, so there's my answer.

00:15:54.350 --> 00:15:56.940
And let's just recap quickly
how we arrived at it.

00:15:56.940 --> 00:16:00.720
So we started with
this integral.

00:16:00.720 --> 00:16:03.859
And we saw, we recognized
it as a good candidate

00:16:03.859 --> 00:16:06.150
for integration by parts,
and so we applied integration

00:16:06.150 --> 00:16:10.350
by parts, seeing that the arctan
part was the part that really

00:16:10.350 --> 00:16:13.837
wanted to be differentiated, and
that the 1 over x squared part,

00:16:13.837 --> 00:16:15.670
you know, it could have
been differentiated,

00:16:15.670 --> 00:16:18.170
it could've been integrated,
but since the arctan

00:16:18.170 --> 00:16:21.600
wanted to be differentiated,
this got integrated.

00:16:24.060 --> 00:16:24.560
Good.

00:16:24.560 --> 00:16:26.520
So we did integration
by parts, and then we're

00:16:26.520 --> 00:16:30.250
left with a rational function
as the piece of the integral we

00:16:30.250 --> 00:16:31.320
had left to compute.

00:16:31.320 --> 00:16:32.880
So when you have a
rational function,

00:16:32.880 --> 00:16:35.140
one method that always
works is that you

00:16:35.140 --> 00:16:36.690
can do partial fractions.

00:16:36.690 --> 00:16:39.380
Now, in this case, this was a
fairly simple partial fractions

00:16:39.380 --> 00:16:39.910
to do.

00:16:39.910 --> 00:16:42.930
First of all, the
degree of the numerator

00:16:42.930 --> 00:16:45.400
was smaller than the
degree of the denominator,

00:16:45.400 --> 00:16:47.320
so we didn't have to
divide by anything,

00:16:47.320 --> 00:16:49.194
you know, we didn't have
to do long division.

00:16:49.194 --> 00:16:52.244
And the denominator had
a fairly simple form,

00:16:52.244 --> 00:16:53.910
so we could just do
the cover up method,

00:16:53.910 --> 00:16:57.384
and then multiply through
and be done fairly quickly.

00:16:57.384 --> 00:16:58.050
But in any case.

00:16:58.050 --> 00:17:00.550
So we did, we applied
partial fractions,

00:17:00.550 --> 00:17:02.110
so we got this expression.

00:17:02.110 --> 00:17:04.180
So then we carried that
back up to our integral,

00:17:04.180 --> 00:17:05.950
and integrated.

00:17:05.950 --> 00:17:09.330
OK, so then we had this integral
with the partial fraction

00:17:09.330 --> 00:17:13.190
expression in it, and that
was easy to integrate.

00:17:13.190 --> 00:17:16.540
So in this case, quite
easy to integrate.

00:17:16.540 --> 00:17:19.240
Sometimes it's a little
messier, but we came out

00:17:19.240 --> 00:17:20.586
pretty luckily this time.

00:17:20.586 --> 00:17:22.320
So OK.

00:17:22.320 --> 00:17:23.170
And so there you go.

00:17:23.170 --> 00:17:25.150
So that's going to
be our final answer.

00:17:25.150 --> 00:17:27.240
Now, this was long
and complicated.

00:17:27.240 --> 00:17:29.610
And so sometimes, when you
do a long and complicated

00:17:29.610 --> 00:17:31.890
computation, you worry, did
I make any stupid mistakes

00:17:31.890 --> 00:17:33.490
along the way?

00:17:33.490 --> 00:17:36.447
One way to check, always, when
you've done an antiderivative,

00:17:36.447 --> 00:17:38.780
is you could always check,
you could take the derivative

00:17:38.780 --> 00:17:39.870
of this expression.

00:17:39.870 --> 00:17:42.070
So that'll require you
to use some product

00:17:42.070 --> 00:17:44.460
rule and some chain rule.

00:17:44.460 --> 00:17:48.320
But, you know,
it's arithmetically

00:17:48.320 --> 00:17:52.760
a little difficult, but not
sort of intellectually, right?

00:17:52.760 --> 00:17:56.470
You just are applying rules
sort of automatically.

00:17:56.470 --> 00:17:58.860
So you can take the
derivative of this expression,

00:17:58.860 --> 00:18:01.940
and check to make sure that it
actually agrees with the thing

00:18:01.940 --> 00:18:04.330
that we started with.

00:18:04.330 --> 00:18:05.820
So good.

00:18:05.820 --> 00:18:07.121
I'll end there.