[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.90,0:00:04.54,Default,,0000,0000,0000,,In this video, we're going to\Nhave a look at how we can
Dialogue: 0,0:00:04.54,0:00:06.72,Default,,0000,0000,0000,,integrate algebraic\Nfractions. The sorts of
Dialogue: 0,0:00:06.72,0:00:09.21,Default,,0000,0000,0000,,fractions that we're going to\Nintegrate and like these
Dialogue: 0,0:00:09.21,0:00:09.49,Default,,0000,0000,0000,,here.
Dialogue: 0,0:00:10.59,0:00:13.99,Default,,0000,0000,0000,,Now, superficially, they will\Nlook very similar, but there are
Dialogue: 0,0:00:13.99,0:00:17.73,Default,,0000,0000,0000,,important differences which I'd\Nlike to point out when you come
Dialogue: 0,0:00:17.73,0:00:20.79,Default,,0000,0000,0000,,to tackle a problem of\Nintegrating a fraction like
Dialogue: 0,0:00:20.79,0:00:24.19,Default,,0000,0000,0000,,this, it's important that you\Ncan look for certain features,
Dialogue: 0,0:00:24.19,0:00:25.89,Default,,0000,0000,0000,,for example, in this first
Dialogue: 0,0:00:25.89,0:00:30.61,Default,,0000,0000,0000,,example. In the denominator we\Nhave what we call 2 linear
Dialogue: 0,0:00:30.61,0:00:33.98,Default,,0000,0000,0000,,factors. These two linear\Nfactors are two different linear
Dialogue: 0,0:00:33.98,0:00:38.09,Default,,0000,0000,0000,,factors. When I say linear\Nfactors, I mean there's no X
Dialogue: 0,0:00:38.09,0:00:42.58,Default,,0000,0000,0000,,squared, no ex cubes, nothing\Nlike that in it. These are just
Dialogue: 0,0:00:42.58,0:00:47.81,Default,,0000,0000,0000,,of the form a X Plus B linear\Nfactors. A constant times X plus
Dialogue: 0,0:00:47.81,0:00:50.81,Default,,0000,0000,0000,,another constant. So with two\Nlinear factors here.
Dialogue: 0,0:00:51.87,0:00:55.96,Default,,0000,0000,0000,,This example is also got linear\Nfactors in clearly X Plus One is
Dialogue: 0,0:00:55.96,0:01:00.72,Default,,0000,0000,0000,,a linear factor. X minus one is\Na linear factor, but the fact
Dialogue: 0,0:01:00.72,0:01:04.44,Default,,0000,0000,0000,,that I've got an X minus one\Nsquared means that we've really
Dialogue: 0,0:01:04.44,0:01:08.47,Default,,0000,0000,0000,,got X minus one times X minus 1\Ntwo linear factors within here.
Dialogue: 0,0:01:08.47,0:01:11.88,Default,,0000,0000,0000,,So we call this an example of a\Nrepeated linear factor.
Dialogue: 0,0:01:13.15,0:01:18.60,Default,,0000,0000,0000,,Linear factors, repeated linear\Nfactors. Over here we've got a
Dialogue: 0,0:01:18.60,0:01:23.50,Default,,0000,0000,0000,,quadratic. Now this particular\Nquadratic will not factorize and
Dialogue: 0,0:01:23.50,0:01:28.96,Default,,0000,0000,0000,,because it won't factorize, we\Ncall it an irreducible quadratic
Dialogue: 0,0:01:28.96,0:01:33.70,Default,,0000,0000,0000,,factor. This final example here\Nhas also got a quadratic factor
Dialogue: 0,0:01:33.70,0:01:37.33,Default,,0000,0000,0000,,in the denominator, but unlike\Nthe previous one, this one will
Dialogue: 0,0:01:37.33,0:01:40.63,Default,,0000,0000,0000,,in fact factorize because X\Nsquared minus four is actually
Dialogue: 0,0:01:40.63,0:01:44.92,Default,,0000,0000,0000,,the difference of two squares\Nand we can write this as X minus
Dialogue: 0,0:01:44.92,0:01:48.88,Default,,0000,0000,0000,,2X plus two. So whilst it might\Nhave originally looked like a
Dialogue: 0,0:01:48.88,0:01:52.51,Default,,0000,0000,0000,,quadratic factor, it was in fact\Nto linear factors, so that's
Dialogue: 0,0:01:52.51,0:01:56.14,Default,,0000,0000,0000,,what those are one of the\Nimportant things we should be
Dialogue: 0,0:01:56.14,0:01:59.44,Default,,0000,0000,0000,,looking for when we come to\Nintegrate quantities like these.
Dialogue: 0,0:01:59.44,0:02:00.76,Default,,0000,0000,0000,,Weather we've got linear
Dialogue: 0,0:02:00.76,0:02:03.54,Default,,0000,0000,0000,,factors. Repeated linear\Nfactors, irreducible quadratic
Dialogue: 0,0:02:03.54,0:02:06.90,Default,,0000,0000,0000,,factors, or quadratic factors\Nthat will factorize.
Dialogue: 0,0:02:07.44,0:02:10.56,Default,,0000,0000,0000,,Something else which is\Nimportant as well, is to examine
Dialogue: 0,0:02:10.56,0:02:14.30,Default,,0000,0000,0000,,the degree of the numerator and\Nthe degree of the denominator in
Dialogue: 0,0:02:14.30,0:02:17.11,Default,,0000,0000,0000,,each of these fractions.\NRemember, the degree is the
Dialogue: 0,0:02:17.11,0:02:20.54,Default,,0000,0000,0000,,highest power, so for example in\Nthe denominator of this example
Dialogue: 0,0:02:20.54,0:02:24.29,Default,,0000,0000,0000,,here, if we multiplied it all\Nout, we actually get the highest
Dialogue: 0,0:02:24.29,0:02:27.72,Default,,0000,0000,0000,,power as three, because when we\Nmultiply the first terms out
Dialogue: 0,0:02:27.72,0:02:31.78,Default,,0000,0000,0000,,will get an X squared, and when\Nwe multiply it with the SEC
Dialogue: 0,0:02:31.78,0:02:36.14,Default,,0000,0000,0000,,bracket, X Plus, one will end up\Nwith an X cubed. So the degree
Dialogue: 0,0:02:36.14,0:02:38.02,Default,,0000,0000,0000,,of the denominator there is 3.
Dialogue: 0,0:02:38.67,0:02:42.95,Default,,0000,0000,0000,,The degree of the numerator is 0\Nbecause we can think of this as
Dialogue: 0,0:02:42.95,0:02:44.48,Default,,0000,0000,0000,,One X to the 0.
Dialogue: 0,0:02:46.04,0:02:50.34,Default,,0000,0000,0000,,In the first case, we've got an\NX to the one here, so the degree
Dialogue: 0,0:02:50.34,0:02:53.79,Default,,0000,0000,0000,,of the numerator there is one,\Nand if we multiply the brackets
Dialogue: 0,0:02:53.79,0:02:57.23,Default,,0000,0000,0000,,out, the degree of the of the\Ndenominator will be two. Will
Dialogue: 0,0:02:57.23,0:03:00.96,Default,,0000,0000,0000,,get a quadratic term in here. In\Nthis case, the degree of the
Dialogue: 0,0:03:00.96,0:03:05.26,Default,,0000,0000,0000,,numerator is 0. And in this\Ncase, the degree of the
Dialogue: 0,0:03:05.26,0:03:09.63,Default,,0000,0000,0000,,denominator is to the highest\Npower is too. So in all of
Dialogue: 0,0:03:09.63,0:03:13.63,Default,,0000,0000,0000,,these cases, the degree of the\Nnumerator is less than the
Dialogue: 0,0:03:13.63,0:03:17.27,Default,,0000,0000,0000,,degree of the denominator, and\Nwe call fractions like these
Dialogue: 0,0:03:17.27,0:03:18.00,Default,,0000,0000,0000,,proper fractions.
Dialogue: 0,0:03:20.11,0:03:24.11,Default,,0000,0000,0000,,On the other hand, if we look at\Nthis final example, the degree
Dialogue: 0,0:03:24.11,0:03:27.81,Default,,0000,0000,0000,,of the numerator is 3, whereas\Nthe degree of the denominator is
Dialogue: 0,0:03:27.81,0:03:31.81,Default,,0000,0000,0000,,too. So in this case, the degree\Nof the numerator is greater than
Dialogue: 0,0:03:31.81,0:03:35.20,Default,,0000,0000,0000,,the degree of the denominator,\Nand this is what's called an
Dialogue: 0,0:03:35.20,0:03:38.67,Default,,0000,0000,0000,,improper fraction. Now when we\Nstart to integrate quantities
Dialogue: 0,0:03:38.67,0:03:41.97,Default,,0000,0000,0000,,like this will need to examine\Nwhether we're dealing with
Dialogue: 0,0:03:41.97,0:03:45.27,Default,,0000,0000,0000,,proper fractions or improper\Nfractions, and then, as I said
Dialogue: 0,0:03:45.27,0:03:49.23,Default,,0000,0000,0000,,before, will need to look at all\Nthe factors in the denominator.
Dialogue: 0,0:03:49.80,0:03:53.48,Default,,0000,0000,0000,,Will also need to call appan\Ntechniques in the theory of
Dialogue: 0,0:03:53.48,0:03:56.84,Default,,0000,0000,0000,,partial fractions. There is a\Nvideo on partial fractions and
Dialogue: 0,0:03:56.84,0:03:59.52,Default,,0000,0000,0000,,you may wish to refer to that if
Dialogue: 0,0:03:59.52,0:04:04.09,Default,,0000,0000,0000,,necessary. If you have a linear\Nfactor in the denominator, this
Dialogue: 0,0:04:04.09,0:04:08.25,Default,,0000,0000,0000,,will lead to a partial fraction\Nof this form. A constant over
Dialogue: 0,0:04:08.25,0:04:09.30,Default,,0000,0000,0000,,the linear factor.
Dialogue: 0,0:04:10.10,0:04:13.74,Default,,0000,0000,0000,,If you have a repeated linear\Nfactor in the denominator,
Dialogue: 0,0:04:13.74,0:04:16.65,Default,,0000,0000,0000,,you'll need two partial\Nfractions. A constant over
Dialogue: 0,0:04:16.65,0:04:19.93,Default,,0000,0000,0000,,the factor and a constant\Nover the factor squared.
Dialogue: 0,0:04:21.84,0:04:25.46,Default,,0000,0000,0000,,Finally, if you have a quadratic\Nfactor which is irreducible,
Dialogue: 0,0:04:25.46,0:04:29.80,Default,,0000,0000,0000,,you'll need to write a partial\Nfraction of the form a constant
Dialogue: 0,0:04:29.80,0:04:33.06,Default,,0000,0000,0000,,times X plus another constant\Nover the irreducible quadratic
Dialogue: 0,0:04:33.06,0:04:36.68,Default,,0000,0000,0000,,factor, so will certainly be\Ncalling upon the techniques of
Dialogue: 0,0:04:36.68,0:04:41.62,Default,,0000,0000,0000,,partial fractions. Will also\Nneed to call appan. Lots of
Dialogue: 0,0:04:41.62,0:04:42.99,Default,,0000,0000,0000,,techniques and integration.
Dialogue: 0,0:04:42.99,0:04:46.55,Default,,0000,0000,0000,,I'm just going to mention just\Ntwo or three here, which will
Dialogue: 0,0:04:46.55,0:04:49.82,Default,,0000,0000,0000,,need to use as we proceed\Nthrough the examples of 1st.
Dialogue: 0,0:04:49.82,0:04:53.09,Default,,0000,0000,0000,,Crucial result is the standard\Nresult, which says that if you
Dialogue: 0,0:04:53.09,0:04:56.06,Default,,0000,0000,0000,,have an integral consisting of a\Nfunction in the denominator.
Dialogue: 0,0:04:56.77,0:05:00.81,Default,,0000,0000,0000,,And it's derivative in the\Nnumerator. Then the result is
Dialogue: 0,0:05:00.81,0:05:05.25,Default,,0000,0000,0000,,the logarithm of the modulus of\Nthe function in the denominator.
Dialogue: 0,0:05:05.25,0:05:10.91,Default,,0000,0000,0000,,So for example, if I ask you to\Nintegrate one over X plus one
Dialogue: 0,0:05:10.91,0:05:12.53,Default,,0000,0000,0000,,with respect to X.
Dialogue: 0,0:05:13.32,0:05:16.82,Default,,0000,0000,0000,,Then clearly the function in the\Ndenominator is X plus one.
Dialogue: 0,0:05:17.39,0:05:20.67,Default,,0000,0000,0000,,And its derivative is one which\Nappears in the numerator. So
Dialogue: 0,0:05:20.67,0:05:22.46,Default,,0000,0000,0000,,we've an example of this form.
Dialogue: 0,0:05:23.64,0:05:27.37,Default,,0000,0000,0000,,So the resulting integral is the\Nlogarithm of the modulus of the
Dialogue: 0,0:05:27.37,0:05:30.48,Default,,0000,0000,0000,,function that was in the\Ndenominator, which is X plus
Dialogue: 0,0:05:30.48,0:05:34.32,Default,,0000,0000,0000,,one. Plus a constant of\Nintegration, so we will need
Dialogue: 0,0:05:34.32,0:05:37.83,Default,,0000,0000,0000,,that result very frequently in\Nthe examples which are going to
Dialogue: 0,0:05:37.83,0:05:41.34,Default,,0000,0000,0000,,follow will also need some\Nstandard results and one of the
Dialogue: 0,0:05:41.34,0:05:44.85,Default,,0000,0000,0000,,standard results I will call\Nappan is this one. The integral
Dialogue: 0,0:05:44.85,0:05:48.100,Default,,0000,0000,0000,,of one over a squared plus X\Nsquared is one over a inverse
Dialogue: 0,0:05:48.100,0:05:51.23,Default,,0000,0000,0000,,tan of X over a plus C.
Dialogue: 0,0:05:52.40,0:05:55.82,Default,,0000,0000,0000,,Results like this can be found\Nin tables of standard integrals.
Dialogue: 0,0:05:57.12,0:06:00.85,Default,,0000,0000,0000,,Finally, we need to integrate\Nquantities like this and you'll
Dialogue: 0,0:06:00.85,0:06:04.58,Default,,0000,0000,0000,,need to do this probably using\Nintegration by substitution. An
Dialogue: 0,0:06:04.58,0:06:08.68,Default,,0000,0000,0000,,integral like this can be worked\Nout by making the substitution
Dialogue: 0,0:06:08.68,0:06:10.55,Default,,0000,0000,0000,,you equals X minus one.
Dialogue: 0,0:06:11.26,0:06:16.60,Default,,0000,0000,0000,,So that the differential du\Nis du DX.
Dialogue: 0,0:06:16.61,0:06:22.79,Default,,0000,0000,0000,,DX, which in this case is du\NDX, will be just one.
Dialogue: 0,0:06:22.85,0:06:27.47,Default,,0000,0000,0000,,So do you is DX and that's\Nintegral. Then will become
Dialogue: 0,0:06:27.47,0:06:30.83,Default,,0000,0000,0000,,the integral of one over\Nyou, squared du.
Dialogue: 0,0:06:32.08,0:06:36.59,Default,,0000,0000,0000,,One over you squared is the same\Nas the integral of you to the
Dialogue: 0,0:06:36.59,0:06:40.13,Default,,0000,0000,0000,,minus 2D U, which you can solve\Nby integrating increasing the
Dialogue: 0,0:06:40.13,0:06:44.64,Default,,0000,0000,0000,,power BI want to give you you to\Nthe minus one over minus one.
Dialogue: 0,0:06:45.29,0:06:46.77,Default,,0000,0000,0000,,Plus a constant of integration.
Dialogue: 0,0:06:48.18,0:06:51.91,Default,,0000,0000,0000,,This can be finished off by\Nchanging the you back to the
Dialogue: 0,0:06:51.91,0:06:55.64,Default,,0000,0000,0000,,original variable X minus one\Nand that will give us X minus
Dialogue: 0,0:06:55.64,0:06:57.82,Default,,0000,0000,0000,,one to the minus one over minus
Dialogue: 0,0:06:57.82,0:07:02.87,Default,,0000,0000,0000,,one plus C. Which is the same as\Nminus one over X minus one plus
Dialogue: 0,0:07:02.87,0:07:04.89,Default,,0000,0000,0000,,C, which is the results I have
Dialogue: 0,0:07:04.89,0:07:09.38,Default,,0000,0000,0000,,here. So what I'm saying is that\Nthroughout the rest of this unit
Dialogue: 0,0:07:09.38,0:07:12.71,Default,,0000,0000,0000,,will need to call Appan lots of\Ndifferent techniques to be able
Dialogue: 0,0:07:12.71,0:07:13.82,Default,,0000,0000,0000,,to perform the integrals.
Dialogue: 0,0:07:14.54,0:07:16.02,Default,,0000,0000,0000,,As we shall see.
Dialogue: 0,0:07:16.59,0:07:19.90,Default,,0000,0000,0000,,Let's look at the
Dialogue: 0,0:07:19.90,0:07:26.49,Default,,0000,0000,0000,,first example. Suppose we\Nwant to integrate this algebraic
Dialogue: 0,0:07:26.49,0:07:29.77,Default,,0000,0000,0000,,fraction. 6 / 2
Dialogue: 0,0:07:29.77,0:07:33.24,Default,,0000,0000,0000,,minus X. X
Dialogue: 0,0:07:33.24,0:07:38.14,Default,,0000,0000,0000,,+3\NDX
Dialogue: 0,0:07:39.30,0:07:42.99,Default,,0000,0000,0000,,The first thing we do is we look\Nat the object we've got and try
Dialogue: 0,0:07:42.99,0:07:45.70,Default,,0000,0000,0000,,to ask ourselves, are we dealing\Nwith a proper or improper
Dialogue: 0,0:07:45.70,0:07:48.16,Default,,0000,0000,0000,,fraction and what are the\Nfactors in the denominator like?
Dialogue: 0,0:07:49.20,0:07:53.21,Default,,0000,0000,0000,,Well, if we multiply the power,\Nthe brackets at the bottom will
Dialogue: 0,0:07:53.21,0:07:57.88,Default,,0000,0000,0000,,find that the highest power of X\Nis X squared, so the degree of
Dialogue: 0,0:07:57.88,0:07:59.22,Default,,0000,0000,0000,,the denominator is 2.
Dialogue: 0,0:08:00.43,0:08:04.32,Default,,0000,0000,0000,,The highest power in the\Nnumerator is one. This is an X
Dialogue: 0,0:08:04.32,0:08:08.53,Default,,0000,0000,0000,,to the power one, so the degree\Nof the numerator is one because
Dialogue: 0,0:08:08.53,0:08:12.42,Default,,0000,0000,0000,,the degree of the numerator is\Nless than the degree of the
Dialogue: 0,0:08:12.42,0:08:16.46,Default,,0000,0000,0000,,denominator. This is an example\Nof a proper fraction.
Dialogue: 0,0:08:17.04,0:08:22.74,Default,,0000,0000,0000,,Both of these factors\Nin the denominator.
Dialogue: 0,0:08:23.38,0:08:24.66,Default,,0000,0000,0000,,Are linear factors.
Dialogue: 0,0:08:25.25,0:08:31.80,Default,,0000,0000,0000,,So we're dealing with a proper\Nfraction with linear factors.
Dialogue: 0,0:08:31.82,0:08:37.79,Default,,0000,0000,0000,,The way we proceed is to take\Nthis fraction and express it in
Dialogue: 0,0:08:37.79,0:08:41.92,Default,,0000,0000,0000,,partial fractions. So I'll start\Nwith the fraction again.
Dialogue: 0,0:08:42.77,0:08:49.04,Default,,0000,0000,0000,,And express it in the\Nappropriate form of partial
Dialogue: 0,0:08:49.04,0:08:51.83,Default,,0000,0000,0000,,fractions. Now because it's
Dialogue: 0,0:08:51.83,0:08:56.21,Default,,0000,0000,0000,,proper. And because we've got\Nlinear factors, the appropriate
Dialogue: 0,0:08:56.21,0:09:00.57,Default,,0000,0000,0000,,form is to have a constant over\Nthe first linear factor.
Dialogue: 0,0:09:01.56,0:09:06.38,Default,,0000,0000,0000,,Plus another constant over the\Nsecond linear factor.
Dialogue: 0,0:09:07.58,0:09:11.53,Default,,0000,0000,0000,,Our task now is to find values\Nfor the constants A&B.
Dialogue: 0,0:09:12.22,0:09:15.98,Default,,0000,0000,0000,,Once we've done that, will\Nbe able to evaluate this
Dialogue: 0,0:09:15.98,0:09:18.24,Default,,0000,0000,0000,,integral by evaluating these\Ntwo separately.
Dialogue: 0,0:09:19.48,0:09:24.14,Default,,0000,0000,0000,,So to find A and be the first\Nthing we do is we add these
Dialogue: 0,0:09:24.14,0:09:25.39,Default,,0000,0000,0000,,two fractions together again.
Dialogue: 0,0:09:26.44,0:09:31.28,Default,,0000,0000,0000,,Remember that to add 2 fractions\Ntogether, we've got to give them
Dialogue: 0,0:09:31.28,0:09:34.90,Default,,0000,0000,0000,,the same denominator. They've\Ngot to have a common
Dialogue: 0,0:09:34.90,0:09:38.53,Default,,0000,0000,0000,,denominator. The common\Ndenominator is going to be made
Dialogue: 0,0:09:38.53,0:09:41.35,Default,,0000,0000,0000,,up of the two factors. 2 minus
Dialogue: 0,0:09:41.35,0:09:46.61,Default,,0000,0000,0000,,X&X +3. To write the first term\Nas an equivalent fraction with
Dialogue: 0,0:09:46.61,0:09:50.68,Default,,0000,0000,0000,,this denominator, we multiply\Ntop and bottom by X plus three.
Dialogue: 0,0:09:50.68,0:09:56.23,Default,,0000,0000,0000,,So if we multiply top here by X\N+3 and bottom there by X +3.
Dialogue: 0,0:09:56.80,0:09:58.52,Default,,0000,0000,0000,,Will achieve this fraction.
Dialogue: 0,0:09:59.21,0:10:01.61,Default,,0000,0000,0000,,And this fraction is equivalent\Nto the original 1.
Dialogue: 0,0:10:03.55,0:10:06.37,Default,,0000,0000,0000,,Similarly with the second term.
Dialogue: 0,0:10:06.96,0:10:12.63,Default,,0000,0000,0000,,To achieve a common denominator\Nof 2 minus XX +3.
Dialogue: 0,0:10:13.31,0:10:15.15,Default,,0000,0000,0000,,I need to multiply top and
Dialogue: 0,0:10:15.15,0:10:20.97,Default,,0000,0000,0000,,bottom here. By two minus X, so\NB times 2 minus X and this
Dialogue: 0,0:10:20.97,0:10:24.81,Default,,0000,0000,0000,,denominator times 2 minus X and\Nthat will give me.
Dialogue: 0,0:10:25.39,0:10:27.31,Default,,0000,0000,0000,,B2 minus X at the top.
Dialogue: 0,0:10:28.42,0:10:32.01,Default,,0000,0000,0000,,Now these two fractions\Nhave the same denominator,
Dialogue: 0,0:10:32.01,0:10:36.05,Default,,0000,0000,0000,,we can add them together\Nsimply by adding the
Dialogue: 0,0:10:36.05,0:10:39.64,Default,,0000,0000,0000,,numerators together, which\Nwill give us a multiplied
Dialogue: 0,0:10:39.64,0:10:40.99,Default,,0000,0000,0000,,by X +3.
Dialogue: 0,0:10:42.13,0:10:47.81,Default,,0000,0000,0000,,Plus B multiplied by\Ntwo minus X.
Dialogue: 0,0:10:48.51,0:10:52.45,Default,,0000,0000,0000,,All divided by the common
Dialogue: 0,0:10:52.45,0:10:59.13,Default,,0000,0000,0000,,denominator. What we're saying\Nis that this fraction we
Dialogue: 0,0:10:59.13,0:11:02.70,Default,,0000,0000,0000,,started with is exactly the
Dialogue: 0,0:11:02.70,0:11:05.20,Default,,0000,0000,0000,,same. As this quantity here.
Dialogue: 0,0:11:06.60,0:11:09.48,Default,,0000,0000,0000,,Now the denominators\Nare already the same.
Dialogue: 0,0:11:11.04,0:11:14.78,Default,,0000,0000,0000,,So if this is the same as that,\Nand the denominators are already
Dialogue: 0,0:11:14.78,0:11:18.53,Default,,0000,0000,0000,,the same, then so too must be\Nthe numerators, so we can equate
Dialogue: 0,0:11:18.53,0:11:21.98,Default,,0000,0000,0000,,the numerators if we equate the\Nnumerators we can write down X
Dialogue: 0,0:11:21.98,0:11:26.75,Default,,0000,0000,0000,,equals. AX\N+3.
Dialogue: 0,0:11:28.28,0:11:32.29,Default,,0000,0000,0000,,Plus B2 Minus\NX.
Dialogue: 0,0:11:33.71,0:11:37.99,Default,,0000,0000,0000,,This is the equation that's\Ngoing to allow us to calculate
Dialogue: 0,0:11:37.99,0:11:39.16,Default,,0000,0000,0000,,values for A&B.
Dialogue: 0,0:11:40.69,0:11:44.12,Default,,0000,0000,0000,,Now we can find values for A&B\Nin one of two different ways.
Dialogue: 0,0:11:44.12,0:11:45.97,Default,,0000,0000,0000,,The 1st way that I'm going to
Dialogue: 0,0:11:45.97,0:11:50.40,Default,,0000,0000,0000,,look at. Is to substitute\Nspecific values in for X.
Dialogue: 0,0:11:51.27,0:11:54.56,Default,,0000,0000,0000,,Remember that this quantity on\Nthe left is supposed to be equal
Dialogue: 0,0:11:54.56,0:11:58.39,Default,,0000,0000,0000,,to this on the right for any\Nvalue of X at all. So in
Dialogue: 0,0:11:58.39,0:12:01.41,Default,,0000,0000,0000,,particular, we can choose any\Nvalues that we like. That will
Dialogue: 0,0:12:01.41,0:12:02.78,Default,,0000,0000,0000,,make all this look simpler.
Dialogue: 0,0:12:03.43,0:12:07.44,Default,,0000,0000,0000,,And what I'm going to do is I'm\Ngoing to choose X to have the
Dialogue: 0,0:12:07.44,0:12:09.30,Default,,0000,0000,0000,,value to. Why would I do that?
Dialogue: 0,0:12:09.82,0:12:13.74,Default,,0000,0000,0000,,I choose X to have the value\Ntoo, because then this second
Dialogue: 0,0:12:13.74,0:12:18.32,Default,,0000,0000,0000,,term will become zero and have 2\N- 2, which is zero. Will lose
Dialogue: 0,0:12:18.32,0:12:20.88,Default,,0000,0000,0000,,this term. And we'll be able to
Dialogue: 0,0:12:20.88,0:12:24.74,Default,,0000,0000,0000,,calculate A. So by careful\Nchoice of values for X, we can
Dialogue: 0,0:12:24.74,0:12:26.25,Default,,0000,0000,0000,,make this look a lot simpler.
Dialogue: 0,0:12:27.60,0:12:28.91,Default,,0000,0000,0000,,So with X is 2.
Dialogue: 0,0:12:29.47,0:12:31.28,Default,,0000,0000,0000,,On the left will have two.
Dialogue: 0,0:12:32.11,0:12:39.08,Default,,0000,0000,0000,,On the right will have 2 +\N3, which is 5 times a.
Dialogue: 0,0:12:39.52,0:12:40.79,Default,,0000,0000,0000,,And this term will vanish.
Dialogue: 0,0:12:41.62,0:12:46.34,Default,,0000,0000,0000,,This gives me a value for a\Nstraightaway dividing both sides
Dialogue: 0,0:12:46.34,0:12:50.20,Default,,0000,0000,0000,,by 5. I can write that a is 2/5.
Dialogue: 0,0:12:52.24,0:12:54.22,Default,,0000,0000,0000,,We need to find B.
Dialogue: 0,0:12:55.66,0:13:00.91,Default,,0000,0000,0000,,Now a sensible value that will\Nenable us to find B is to let X
Dialogue: 0,0:13:00.91,0:13:03.01,Default,,0000,0000,0000,,be minus three whi, is that?
Dialogue: 0,0:13:03.53,0:13:06.77,Default,,0000,0000,0000,,Well, if X is minus three,\Nwill have minus 3 + 3, which
Dialogue: 0,0:13:06.77,0:13:09.26,Default,,0000,0000,0000,,is zero. And all of this\Nfirst term will vanish.
Dialogue: 0,0:13:10.97,0:13:17.41,Default,,0000,0000,0000,,And we'll be able to find be so\Nletting XP minus three will have
Dialogue: 0,0:13:17.41,0:13:22.93,Default,,0000,0000,0000,,minus three on the left zero\Nfrom this term here, and two
Dialogue: 0,0:13:22.93,0:13:27.99,Default,,0000,0000,0000,,minus minus three, which is 2 +\N3, which is 55-B.
Dialogue: 0,0:13:28.23,0:13:35.84,Default,,0000,0000,0000,,Dividing both sides by 5 will\Ngive us a value for B's
Dialogue: 0,0:13:35.84,0:13:38.37,Default,,0000,0000,0000,,minus three over 5.
Dialogue: 0,0:13:39.28,0:13:43.24,Default,,0000,0000,0000,,So now we know a value for a. We\Nknow a value for B.
Dialogue: 0,0:13:44.25,0:13:47.61,Default,,0000,0000,0000,,And we can then proceed to\Nevaluate the integral by
Dialogue: 0,0:13:47.61,0:13:48.95,Default,,0000,0000,0000,,evaluating each of these
Dialogue: 0,0:13:48.95,0:13:56.03,Default,,0000,0000,0000,,separately. Let me write\Nthis down again. We want
Dialogue: 0,0:13:56.03,0:14:03.56,Default,,0000,0000,0000,,the integral of X divided\Nby 2 minus XX +3.
Dialogue: 0,0:14:03.60,0:14:06.26,Default,,0000,0000,0000,,With respect to X.
Dialogue: 0,0:14:06.26,0:14:09.96,Default,,0000,0000,0000,,We expressed this algebraic\Nfraction in its partial fraction
Dialogue: 0,0:14:09.96,0:14:14.48,Default,,0000,0000,0000,,in its partial fractions, and we\Nfound that a was 2/5.
Dialogue: 0,0:14:15.17,0:14:18.82,Default,,0000,0000,0000,,And be was
Dialogue: 0,0:14:18.82,0:14:25.02,Default,,0000,0000,0000,,minus 3/5. So instead\Nof integrating this original
Dialogue: 0,0:14:25.02,0:14:31.39,Default,,0000,0000,0000,,fraction, what we're going to do\Nnow is integrate separately the
Dialogue: 0,0:14:31.39,0:14:33.13,Default,,0000,0000,0000,,two partial fractions.
Dialogue: 0,0:14:33.25,0:14:38.08,Default,,0000,0000,0000,,And will integrate these\Nseparately and will do it like
Dialogue: 0,0:14:38.08,0:14:42.09,Default,,0000,0000,0000,,this. In the first integral,\Nwe're going to take out the
Dialogue: 0,0:14:42.09,0:14:47.31,Default,,0000,0000,0000,,factor of 2/5. I will be left\Nwith the problem of integrating
Dialogue: 0,0:14:47.31,0:14:50.23,Default,,0000,0000,0000,,one over 2 minus X with respect
Dialogue: 0,0:14:50.23,0:14:57.05,Default,,0000,0000,0000,,to X. For the second,\Nwe're going to take out
Dialogue: 0,0:14:57.05,0:15:03.26,Default,,0000,0000,0000,,minus 3/5. And integrate\None over X +3 with
Dialogue: 0,0:15:03.26,0:15:05.34,Default,,0000,0000,0000,,respect to X.
Dialogue: 0,0:15:07.71,0:15:09.89,Default,,0000,0000,0000,,So the problem of\Nintegrating this algebraic
Dialogue: 0,0:15:09.89,0:15:12.38,Default,,0000,0000,0000,,fraction has been split\Ninto the problem of
Dialogue: 0,0:15:12.38,0:15:14.55,Default,,0000,0000,0000,,evaluating these two\Nseparate integrals and both
Dialogue: 0,0:15:14.55,0:15:17.66,Default,,0000,0000,0000,,of these are simpler than\Nthe one we started with.
Dialogue: 0,0:15:18.70,0:15:22.60,Default,,0000,0000,0000,,Let's deal with the second one\Nfirst. The second one is a
Dialogue: 0,0:15:22.60,0:15:26.18,Default,,0000,0000,0000,,situation where we've got a\Nfunction at the bottom and it's
Dialogue: 0,0:15:26.18,0:15:30.08,Default,,0000,0000,0000,,derivative at the top. Because\Nwe've got X plus three at the
Dialogue: 0,0:15:30.08,0:15:34.30,Default,,0000,0000,0000,,bottom and the derivative of X\N+3 is just one which appears at
Dialogue: 0,0:15:34.30,0:15:37.88,Default,,0000,0000,0000,,the top. So this just evaluates\Nto minus 3/5 the natural
Dialogue: 0,0:15:37.88,0:15:40.80,Default,,0000,0000,0000,,logarithm of the modulus of\Nwhat's at the bottom.
Dialogue: 0,0:15:42.57,0:15:45.47,Default,,0000,0000,0000,,We've got the similar situation\Nhere, except if you
Dialogue: 0,0:15:45.47,0:15:48.37,Default,,0000,0000,0000,,differentiate the denominator,\Nyou get minus one because of
Dialogue: 0,0:15:48.37,0:15:52.55,Default,,0000,0000,0000,,this minus X, so we'd really\Nlike a minus one at the top.
Dialogue: 0,0:15:53.62,0:15:57.61,Default,,0000,0000,0000,,And I can adjust my numerator to\Nmake it minus one, provided that
Dialogue: 0,0:15:57.61,0:16:00.68,Default,,0000,0000,0000,,I counteract that with putting a\Nminus sign outside there.
Dialogue: 0,0:16:01.67,0:16:06.60,Default,,0000,0000,0000,,So we can write all this as\Nminus 2/5 the natural logarithm
Dialogue: 0,0:16:06.60,0:16:12.77,Default,,0000,0000,0000,,of the modulus of 2 minus X. And\Nof course we need a constant of
Dialogue: 0,0:16:12.77,0:16:14.82,Default,,0000,0000,0000,,integration at the very end.
Dialogue: 0,0:16:15.47,0:16:20.53,Default,,0000,0000,0000,,So that's the result of\Nintegrating X over 2 minus XX +3
Dialogue: 0,0:16:20.53,0:16:22.22,Default,,0000,0000,0000,,and the problems finished.
Dialogue: 0,0:16:22.92,0:16:28.42,Default,,0000,0000,0000,,What I'd like to do is just go\Nback a page and just show you an
Dialogue: 0,0:16:28.42,0:16:31.86,Default,,0000,0000,0000,,alternative way of calculating\Nvalues for the constants A&B in
Dialogue: 0,0:16:31.86,0:16:35.65,Default,,0000,0000,0000,,the partial fractions, and I\Nwant us to return to this
Dialogue: 0,0:16:35.65,0:16:39.78,Default,,0000,0000,0000,,equation here that we use to\Nfind A&B. Let me write that
Dialogue: 0,0:16:39.78,0:16:40.81,Default,,0000,0000,0000,,equation down again.
Dialogue: 0,0:16:41.49,0:16:47.58,Default,,0000,0000,0000,,X is equal to\Na X +3.
Dialogue: 0,0:16:47.58,0:16:50.70,Default,,0000,0000,0000,,Plus B2 Minus
Dialogue: 0,0:16:50.70,0:16:58.17,Default,,0000,0000,0000,,X. What I'm going\Nto do is I'm going to
Dialogue: 0,0:16:58.17,0:17:01.42,Default,,0000,0000,0000,,start by removing the brackets.
Dialogue: 0,0:17:01.65,0:17:05.31,Default,,0000,0000,0000,,Will have a multiplied by X
Dialogue: 0,0:17:05.31,0:17:10.15,Default,,0000,0000,0000,,AX. A Times 3\Nwhich is 3A.
Dialogue: 0,0:17:11.94,0:17:15.26,Default,,0000,0000,0000,,B times two or two B.
Dialogue: 0,0:17:16.13,0:17:19.35,Default,,0000,0000,0000,,And be times minus X or\Nminus BX.
Dialogue: 0,0:17:21.05,0:17:24.77,Default,,0000,0000,0000,,And then what I'm going to do\Nis, I'm going to collect similar
Dialogue: 0,0:17:24.77,0:17:28.20,Default,,0000,0000,0000,,terms together so you see Ivan\NAxe here and minus BX there.
Dialogue: 0,0:17:28.77,0:17:33.82,Default,,0000,0000,0000,,So altogether I have a minus B,\Nlots of X.
Dialogue: 0,0:17:37.28,0:17:41.74,Default,,0000,0000,0000,,And we've got 3A Plus\N2B here.
Dialogue: 0,0:17:44.62,0:17:47.89,Default,,0000,0000,0000,,We now use this equation to\Nequate coefficients on both
Dialogue: 0,0:17:47.89,0:17:52.47,Default,,0000,0000,0000,,sides. What do we mean by that?\NWell, what we do is we ask
Dialogue: 0,0:17:52.47,0:17:57.05,Default,,0000,0000,0000,,ourselves how many X terms do we\Nhave on the left and match that
Dialogue: 0,0:17:57.05,0:18:01.62,Default,,0000,0000,0000,,with the number of X terms that\Nwe have on the right. So you
Dialogue: 0,0:18:01.62,0:18:06.53,Default,,0000,0000,0000,,see, on the left hand side here,\Nif we look at just the ex terms,
Dialogue: 0,0:18:06.53,0:18:13.23,Default,,0000,0000,0000,,there's 1X. On the right we've\Ngot a minus B, lots of X.
Dialogue: 0,0:18:13.26,0:18:18.06,Default,,0000,0000,0000,,So we've equated the\Ncoefficients of X on both sides.
Dialogue: 0,0:18:19.36,0:18:22.95,Default,,0000,0000,0000,,We can also look at constant\Nterms on both sides. You see the
Dialogue: 0,0:18:22.95,0:18:24.88,Default,,0000,0000,0000,,three A plus 2B is a constant.
Dialogue: 0,0:18:25.68,0:18:30.36,Default,,0000,0000,0000,,There are no constant terms on\Nthe left, so if we just look at
Dialogue: 0,0:18:30.36,0:18:32.36,Default,,0000,0000,0000,,constants, there are none on the
Dialogue: 0,0:18:32.36,0:18:37.36,Default,,0000,0000,0000,,left. And on the right there's\N3A Plus 2B.
Dialogue: 0,0:18:37.36,0:18:40.69,Default,,0000,0000,0000,,And you'll see what we have.\NHere are two simultaneous
Dialogue: 0,0:18:40.69,0:18:44.35,Default,,0000,0000,0000,,equations for A&B and if we\Nsolve these equations we can
Dialogue: 0,0:18:44.35,0:18:48.68,Default,,0000,0000,0000,,find values for A&B. Let me call\Nthat equation one and that one
Dialogue: 0,0:18:48.68,0:18:52.68,Default,,0000,0000,0000,,equation two. What I'm going to\Ndo is I'm going to multiply
Dialogue: 0,0:18:52.68,0:18:57.67,Default,,0000,0000,0000,,equation one by two so that will\Nend up with two be so that we
Dialogue: 0,0:18:57.67,0:19:01.34,Default,,0000,0000,0000,,will be able to add these\Ntogether to eliminate the bees.
Dialogue: 0,0:19:01.34,0:19:06.33,Default,,0000,0000,0000,,So if I take equation one and I\Nmultiply it by two, I'll get 2
Dialogue: 0,0:19:06.33,0:19:13.54,Default,,0000,0000,0000,,ones or two. 2A minus 2B.\NLet's call that equation 3.
Dialogue: 0,0:19:14.35,0:19:18.03,Default,,0000,0000,0000,,If we add equations two and
Dialogue: 0,0:19:18.03,0:19:24.62,Default,,0000,0000,0000,,three together. We've got 0 + 2,\Nwhich is 2, three, 8 + 2 A which
Dialogue: 0,0:19:24.62,0:19:30.75,Default,,0000,0000,0000,,is 5A and two be added to minus\N2B cancels out, so two is 5. In
Dialogue: 0,0:19:30.75,0:19:35.73,Default,,0000,0000,0000,,other words, A is 2 over 5,\Nwhich is the value we had
Dialogue: 0,0:19:35.73,0:19:37.26,Default,,0000,0000,0000,,earlier on for A.
Dialogue: 0,0:19:37.82,0:19:42.58,Default,,0000,0000,0000,,We can then take this value for\NA and substitute it in either of
Dialogue: 0,0:19:42.58,0:19:46.32,Default,,0000,0000,0000,,these equations and obtain a\Nvalue for be. So, for example,
Dialogue: 0,0:19:46.32,0:19:48.36,Default,,0000,0000,0000,,if we substitute in the first
Dialogue: 0,0:19:48.36,0:19:54.79,Default,,0000,0000,0000,,equation. Will find that one\Nequals a, is 2/5 minus B.
Dialogue: 0,0:19:56.10,0:20:02.05,Default,,0000,0000,0000,,Rearranging this B is equal to\N2/5 - 1.
Dialogue: 0,0:20:02.79,0:20:08.78,Default,,0000,0000,0000,,And 2/5 - 1 is minus 3/5 the\Nsame value as we got before.
Dialogue: 0,0:20:09.76,0:20:13.50,Default,,0000,0000,0000,,So we've seen two ways of\Nfinding the values of the
Dialogue: 0,0:20:13.50,0:20:17.58,Default,,0000,0000,0000,,constants A&B. We can substitute\Nspecific values for X or we can
Dialogue: 0,0:20:17.58,0:20:18.94,Default,,0000,0000,0000,,equate coefficients on both
Dialogue: 0,0:20:18.94,0:20:23.21,Default,,0000,0000,0000,,sides. Often will need to use a\Nmix of the two methods in order
Dialogue: 0,0:20:23.21,0:20:24.92,Default,,0000,0000,0000,,to find all the constants in a
Dialogue: 0,0:20:24.92,0:20:31.40,Default,,0000,0000,0000,,given problem. Let's have\Na look at
Dialogue: 0,0:20:31.40,0:20:34.46,Default,,0000,0000,0000,,a definite integral.
Dialogue: 0,0:20:35.17,0:20:39.03,Default,,0000,0000,0000,,Suppose we want to find the\Nintegral from X is one to access
Dialogue: 0,0:20:39.03,0:20:46.56,Default,,0000,0000,0000,,2. Of three divided by\NXX plus one with respect to
Dialogue: 0,0:20:46.56,0:20:52.80,Default,,0000,0000,0000,,X. As before, we examine this\Nintegrand and ask ourselves, is
Dialogue: 0,0:20:52.80,0:20:55.18,Default,,0000,0000,0000,,this a proper or improper
Dialogue: 0,0:20:55.18,0:20:59.04,Default,,0000,0000,0000,,fraction? Well, the degree of\Nthe denominator is too, because
Dialogue: 0,0:20:59.04,0:21:03.01,Default,,0000,0000,0000,,when we multiply this out, the\Nhighest power of X will be 2.
Dialogue: 0,0:21:03.62,0:21:09.24,Default,,0000,0000,0000,,The degree of the numerator is\Nzero, with really 3X to the zero
Dialogue: 0,0:21:09.24,0:21:12.69,Default,,0000,0000,0000,,here, so this is an example of a
Dialogue: 0,0:21:12.69,0:21:18.09,Default,,0000,0000,0000,,proper fraction. On both of\Nthese factors are linear
Dialogue: 0,0:21:18.09,0:21:23.96,Default,,0000,0000,0000,,factors. So as before, I'm going\Nto express the integrand.
Dialogue: 0,0:21:24.58,0:21:27.04,Default,,0000,0000,0000,,As the sum of its partial\Nfractions. So let's do that
Dialogue: 0,0:21:27.04,0:21:31.04,Default,,0000,0000,0000,,first of all. 3 divided by XX
Dialogue: 0,0:21:31.04,0:21:35.96,Default,,0000,0000,0000,,plus one. The appropriate form\Nof partial fractions.
Dialogue: 0,0:21:36.93,0:21:42.75,Default,,0000,0000,0000,,Are constant. Over the\Nfirst linear factor.
Dialogue: 0,0:21:42.75,0:21:47.52,Default,,0000,0000,0000,,Plus another constant over the\Nsecond linear factor.
Dialogue: 0,0:21:49.16,0:21:52.65,Default,,0000,0000,0000,,And our job now is to try to\Nfind values for A&B.
Dialogue: 0,0:21:53.72,0:22:00.12,Default,,0000,0000,0000,,We do this by adding these\Ntogether as we did before,
Dialogue: 0,0:22:00.12,0:22:04.78,Default,,0000,0000,0000,,common denominator XX plus one\Nin both cases.
Dialogue: 0,0:22:06.81,0:22:10.87,Default,,0000,0000,0000,,To write a over X as an\Nequivalent fraction with
Dialogue: 0,0:22:10.87,0:22:14.12,Default,,0000,0000,0000,,this denominator will\Nneed to multiply top and
Dialogue: 0,0:22:14.12,0:22:16.15,Default,,0000,0000,0000,,bottom by X plus one.
Dialogue: 0,0:22:18.09,0:22:23.93,Default,,0000,0000,0000,,To write B over X plus one with\Nthis denominator will need to
Dialogue: 0,0:22:23.93,0:22:26.62,Default,,0000,0000,0000,,multiply top and bottom by X.
Dialogue: 0,0:22:27.42,0:22:32.84,Default,,0000,0000,0000,,So now we've given these two\Nfractions a common denominator,
Dialogue: 0,0:22:32.84,0:22:37.72,Default,,0000,0000,0000,,and we add the fractions\Ntogether by adding the
Dialogue: 0,0:22:37.72,0:22:43.30,Default,,0000,0000,0000,,numerators. I'm putting the\Nresult over the common
Dialogue: 0,0:22:43.30,0:22:51.22,Default,,0000,0000,0000,,denominator. So 3 divided by XX\Nplus One is equal to all this.
Dialogue: 0,0:22:53.15,0:22:54.80,Default,,0000,0000,0000,,The denominators are already the
Dialogue: 0,0:22:54.80,0:23:00.05,Default,,0000,0000,0000,,same. So we can equate the\Nnumerators that gives us the
Dialogue: 0,0:23:00.05,0:23:05.43,Default,,0000,0000,0000,,equation 3 equals a X plus one\Nplus BX. And this is the
Dialogue: 0,0:23:05.43,0:23:09.99,Default,,0000,0000,0000,,equation we can use to try to\Nfind values for A&B.
Dialogue: 0,0:23:11.26,0:23:13.64,Default,,0000,0000,0000,,We could equate coefficients, or\Nwe can substitute specific
Dialogue: 0,0:23:13.64,0:23:17.33,Default,,0000,0000,0000,,values for X and what I'm going\Nto do is I'm going to substitute
Dialogue: 0,0:23:17.33,0:23:21.03,Default,,0000,0000,0000,,the value X is not and the\Nreason why I'm picking X is not
Dialogue: 0,0:23:21.03,0:23:23.40,Default,,0000,0000,0000,,is because I recognize straight\Naway that's going to.
Dialogue: 0,0:23:23.96,0:23:26.55,Default,,0000,0000,0000,,Kill off this last term here\Nthat'll have gone and will be
Dialogue: 0,0:23:26.55,0:23:28.28,Default,,0000,0000,0000,,able to just find a value for A.
Dialogue: 0,0:23:29.28,0:23:34.92,Default,,0000,0000,0000,,So we substitute X is not on the\Nleft, will still have 3.
Dialogue: 0,0:23:35.79,0:23:41.60,Default,,0000,0000,0000,,And on the right we've got not\Nplus one which is one 1A.
Dialogue: 0,0:23:41.63,0:23:43.61,Default,,0000,0000,0000,,Be times not is not so that
Dialogue: 0,0:23:43.61,0:23:48.25,Default,,0000,0000,0000,,goes. So In other words, we've\Ngot a value for A and a is 3.
Dialogue: 0,0:23:49.64,0:23:54.32,Default,,0000,0000,0000,,Another sensible value to\Nsubstitute is X equals minus
Dialogue: 0,0:23:54.32,0:23:56.56,Default,,0000,0000,0000,,one. Why is that a sensible
Dialogue: 0,0:23:56.56,0:24:00.29,Default,,0000,0000,0000,,value? Well, that's a sensible\Nvalue, because if we put X is
Dialogue: 0,0:24:00.29,0:24:03.71,Default,,0000,0000,0000,,minus one in minus one plus one\Nis zero and will lose this first
Dialogue: 0,0:24:03.71,0:24:05.66,Default,,0000,0000,0000,,term with the A in and will now
Dialogue: 0,0:24:05.66,0:24:09.81,Default,,0000,0000,0000,,be. So putting X is minus one\Nwill have 3.
Dialogue: 0,0:24:10.36,0:24:14.13,Default,,0000,0000,0000,,This will become zero and will\Nhave be times minus one which is
Dialogue: 0,0:24:14.13,0:24:19.70,Default,,0000,0000,0000,,minus B. So this tells us that B\Nis actually minus three.
Dialogue: 0,0:24:20.39,0:24:26.37,Default,,0000,0000,0000,,So now we know the value of a is\Ngoing to be 3 and B is going to
Dialogue: 0,0:24:26.37,0:24:30.88,Default,,0000,0000,0000,,be minus three. And the problem\Nof performing this integration
Dialogue: 0,0:24:30.88,0:24:34.24,Default,,0000,0000,0000,,can be solved by integrating\Nthese two terms separately.
Dialogue: 0,0:24:35.63,0:24:41.31,Default,,0000,0000,0000,,Let's do that now. I'll write\Nthese these terms down again.
Dialogue: 0,0:24:41.31,0:24:46.98,Default,,0000,0000,0000,,We're integrating three over XX\Nplus one with respect to X.
Dialogue: 0,0:24:48.18,0:24:52.31,Default,,0000,0000,0000,,And we've expressed already this\Nas its partial fractions, and
Dialogue: 0,0:24:52.31,0:24:56.85,Default,,0000,0000,0000,,found that we're integrating\Nthree over X minus three over X,
Dialogue: 0,0:24:56.85,0:25:02.22,Default,,0000,0000,0000,,plus one with respect to X, and\Nthis was a definite integral. It
Dialogue: 0,0:25:02.22,0:25:06.35,Default,,0000,0000,0000,,had limits on, and the limits\Nwill one and two.
Dialogue: 0,0:25:06.89,0:25:10.26,Default,,0000,0000,0000,,So now we use partial\Nfractions to change this
Dialogue: 0,0:25:10.26,0:25:12.87,Default,,0000,0000,0000,,algebraic fraction into\Nthese two simple integrals.
Dialogue: 0,0:25:13.97,0:25:16.09,Default,,0000,0000,0000,,Now, these are\Nstraightforward to finish
Dialogue: 0,0:25:16.09,0:25:19.62,Default,,0000,0000,0000,,because the integral of\Nthree over X is just three
Dialogue: 0,0:25:19.62,0:25:22.09,Default,,0000,0000,0000,,natural logarithm of the\Nmodulus of X.
Dialogue: 0,0:25:25.65,0:25:27.41,Default,,0000,0000,0000,,The integral of three over X
Dialogue: 0,0:25:27.41,0:25:32.74,Default,,0000,0000,0000,,plus one. Is 3 natural logarithm\Nof the modulus of X plus one?
Dialogue: 0,0:25:33.46,0:25:35.80,Default,,0000,0000,0000,,And there's a minus sign\Nin the middle from that.
Dialogue: 0,0:25:36.86,0:25:40.18,Default,,0000,0000,0000,,This is a definite
Dialogue: 0,0:25:40.18,0:25:44.11,Default,,0000,0000,0000,,integral. So we have square\Nbrackets and we write the limits
Dialogue: 0,0:25:44.11,0:25:45.66,Default,,0000,0000,0000,,on the right hand side.
Dialogue: 0,0:25:46.35,0:25:48.89,Default,,0000,0000,0000,,The problem is nearly\Nfinished. All we have to do
Dialogue: 0,0:25:48.89,0:25:50.16,Default,,0000,0000,0000,,is substitute the limits in.
Dialogue: 0,0:25:51.31,0:25:56.59,Default,,0000,0000,0000,,Upper limit first when X is\N2, will have three natural
Dialogue: 0,0:25:56.59,0:25:58.03,Default,,0000,0000,0000,,log of two.
Dialogue: 0,0:25:59.15,0:26:02.92,Default,,0000,0000,0000,,When X is 2 in here will have
Dialogue: 0,0:26:02.92,0:26:06.48,Default,,0000,0000,0000,,minus three. Natural\Nlogarithm of 2 + 1, which is
Dialogue: 0,0:26:06.48,0:26:10.73,Default,,0000,0000,0000,,3. So that's what we get when\Nwe put the upper limit in.
Dialogue: 0,0:26:11.75,0:26:16.52,Default,,0000,0000,0000,,When we put the lower limit in,\Nwhen X is one will have three
Dialogue: 0,0:26:16.52,0:26:17.89,Default,,0000,0000,0000,,natural logarithm of 1.
Dialogue: 0,0:26:18.64,0:26:23.40,Default,,0000,0000,0000,,Minus three natural logarithm of\N1 + 1, which is 2. So that's
Dialogue: 0,0:26:23.40,0:26:28.16,Default,,0000,0000,0000,,what we get when we put the\Nlower limiting. And of course we
Dialogue: 0,0:26:28.16,0:26:31.45,Default,,0000,0000,0000,,want to find the difference of\Nthese two quantities.
Dialogue: 0,0:26:32.30,0:26:36.12,Default,,0000,0000,0000,,Here you'll notice with three
Dialogue: 0,0:26:36.12,0:26:40.63,Default,,0000,0000,0000,,log 2. And over here\Nthere's another three
Dialogue: 0,0:26:40.63,0:26:44.27,Default,,0000,0000,0000,,log, two with A minus and\Nminus, so we're adding
Dialogue: 0,0:26:44.27,0:26:47.55,Default,,0000,0000,0000,,another three log 2. So\Naltogether there will be
Dialogue: 0,0:26:47.55,0:26:48.64,Default,,0000,0000,0000,,6 log 2.
Dialogue: 0,0:26:49.84,0:26:53.22,Default,,0000,0000,0000,,That's minus three log 3.
Dialogue: 0,0:26:53.22,0:26:56.67,Default,,0000,0000,0000,,And the logarithm of
Dialogue: 0,0:26:56.67,0:27:00.54,Default,,0000,0000,0000,,1. Is 0 so that banishes.
Dialogue: 0,0:27:01.43,0:27:04.46,Default,,0000,0000,0000,,Now we could leave the answer\Nlike that, although more often
Dialogue: 0,0:27:04.46,0:27:07.48,Default,,0000,0000,0000,,than not would probably use\Nthe laws of logarithms to try
Dialogue: 0,0:27:07.48,0:27:10.78,Default,,0000,0000,0000,,to tighten this up a little\Nbit and write it in a
Dialogue: 0,0:27:10.78,0:27:13.26,Default,,0000,0000,0000,,different way. You should be\Naware that multiplier outside,
Dialogue: 0,0:27:13.26,0:27:16.83,Default,,0000,0000,0000,,like this six, can be put\Ninside as a power, so we can
Dialogue: 0,0:27:16.83,0:27:19.58,Default,,0000,0000,0000,,write this as logarithm of 2\Nto the power 6.
Dialogue: 0,0:27:21.04,0:27:23.96,Default,,0000,0000,0000,,Subtract again a multiplier\Noutside can move inside as a
Dialogue: 0,0:27:23.96,0:27:28.05,Default,,0000,0000,0000,,power so we can write this as\Nlogarithm of 3 to the power 3.
Dialogue: 0,0:27:29.31,0:27:32.97,Default,,0000,0000,0000,,And you'll also be aware from\Nyour loss of logarithms that if
Dialogue: 0,0:27:32.97,0:27:36.32,Default,,0000,0000,0000,,we're finding the difference of\Ntwo logarithms, and we can write
Dialogue: 0,0:27:36.32,0:27:41.20,Default,,0000,0000,0000,,that as the logarithm of 2 to\Nthe power 6 / 3 to the power 3.
Dialogue: 0,0:27:41.45,0:27:44.03,Default,,0000,0000,0000,,And that's my final answer.
Dialogue: 0,0:27:45.28,0:27:52.31,Default,,0000,0000,0000,,Let's look at another example in\Nwhich the denominator contains a
Dialogue: 0,0:27:52.31,0:27:57.42,Default,,0000,0000,0000,,repeated linear factor. Suppose\Nwe're interested in evaluating
Dialogue: 0,0:27:57.42,0:28:05.09,Default,,0000,0000,0000,,this integral 1 divided by X\Nminus one all squared X Plus
Dialogue: 0,0:28:05.09,0:28:12.12,Default,,0000,0000,0000,,One, and we want to integrate\Nthat with respect to X.
Dialogue: 0,0:28:12.99,0:28:14.86,Default,,0000,0000,0000,,So again, we have a proper
Dialogue: 0,0:28:14.86,0:28:19.44,Default,,0000,0000,0000,,fraction. And there is a linear\Nfactor here. X plus one, another
Dialogue: 0,0:28:19.44,0:28:23.48,Default,,0000,0000,0000,,linear factor X minus one. But\Nthis is a repeated linear factor
Dialogue: 0,0:28:23.48,0:28:24.83,Default,,0000,0000,0000,,because it occurs twice.
Dialogue: 0,0:28:25.53,0:28:30.80,Default,,0000,0000,0000,,The appropriate form of partial\Nfractions will be these.
Dialogue: 0,0:28:31.47,0:28:37.61,Default,,0000,0000,0000,,We want to\Nconstant over the
Dialogue: 0,0:28:37.61,0:28:40.68,Default,,0000,0000,0000,,linear factor X
Dialogue: 0,0:28:40.68,0:28:47.19,Default,,0000,0000,0000,,minus one. We want\Nanother constant over the linear
Dialogue: 0,0:28:47.19,0:28:52.69,Default,,0000,0000,0000,,factor repeated X minus one\Nsquared. And finally we need
Dialogue: 0,0:28:52.69,0:28:58.19,Default,,0000,0000,0000,,another constant. See over this\Nlinear factor X plus one.
Dialogue: 0,0:28:58.19,0:29:02.04,Default,,0000,0000,0000,,And our task is before\Nis to try to find values
Dialogue: 0,0:29:02.04,0:29:03.44,Default,,0000,0000,0000,,for the constants AB&C.
Dialogue: 0,0:29:04.47,0:29:09.04,Default,,0000,0000,0000,,We do that as before, by\Nexpressing each of these over a
Dialogue: 0,0:29:09.04,0:29:12.47,Default,,0000,0000,0000,,common denominator and the\Ncommon denominator that we want
Dialogue: 0,0:29:12.47,0:29:15.14,Default,,0000,0000,0000,,is going to be X minus one.
Dialogue: 0,0:29:15.70,0:29:19.74,Default,,0000,0000,0000,,Squared X plus\None.
Dialogue: 0,0:29:20.78,0:29:24.56,Default,,0000,0000,0000,,Now to achieve a common\Ndenominator of X minus one
Dialogue: 0,0:29:24.56,0:29:29.47,Default,,0000,0000,0000,,squared X Plus one will need to\Nmultiply the top and bottom here
Dialogue: 0,0:29:29.47,0:29:35.14,Default,,0000,0000,0000,,by X minus 1X plus one. So we\Nhave a X minus 1X plus one.
Dialogue: 0,0:29:35.17,0:29:38.46,Default,,0000,0000,0000,,To achieve the common\Ndenominator in this case will
Dialogue: 0,0:29:38.46,0:29:42.10,Default,,0000,0000,0000,,need to multiply top and bottom\Nby X plus one.
Dialogue: 0,0:29:42.62,0:29:46.55,Default,,0000,0000,0000,,So we'll have a BX plus one.
Dialogue: 0,0:29:47.12,0:29:51.15,Default,,0000,0000,0000,,And finally, in this case, to\Nachieve a denominator of X minus
Dialogue: 0,0:29:51.15,0:29:55.86,Default,,0000,0000,0000,,one squared X Plus one will need\Nto multiply top and bottom by X
Dialogue: 0,0:29:55.86,0:29:56.86,Default,,0000,0000,0000,,minus 1 squared.
Dialogue: 0,0:29:57.64,0:30:04.38,Default,,0000,0000,0000,,Now this fraction here\Nis the same as
Dialogue: 0,0:30:04.38,0:30:06.90,Default,,0000,0000,0000,,this fraction here.
Dialogue: 0,0:30:07.30,0:30:10.48,Default,,0000,0000,0000,,Their denominators are already\Nthe same, so we can equate the
Dialogue: 0,0:30:10.48,0:30:13.95,Default,,0000,0000,0000,,numerators. So if we just look\Nat the numerators will have one
Dialogue: 0,0:30:13.95,0:30:17.99,Default,,0000,0000,0000,,on the left is equal to the top\Nline here on the right hand
Dialogue: 0,0:30:17.99,0:30:25.16,Default,,0000,0000,0000,,side. AX minus\N1X plus one.
Dialogue: 0,0:30:25.29,0:30:29.21,Default,,0000,0000,0000,,Plus B. X plus one.
Dialogue: 0,0:30:30.29,0:30:34.11,Default,,0000,0000,0000,,Plus C. X minus one all
Dialogue: 0,0:30:34.11,0:30:38.72,Default,,0000,0000,0000,,squared. And now we choose some\Nsensible values for X, so
Dialogue: 0,0:30:38.72,0:30:42.43,Default,,0000,0000,0000,,there's a lot of these terms\Nwill drop away. For example,
Dialogue: 0,0:30:42.43,0:30:46.47,Default,,0000,0000,0000,,supposing we pick X equals 1,\Nwhat's the point of picking X
Dialogue: 0,0:30:46.47,0:30:50.18,Default,,0000,0000,0000,,equals one? Well, if we pick\Naxes one, this first term
Dialogue: 0,0:30:50.18,0:30:51.53,Default,,0000,0000,0000,,vanishes. We lose a.
Dialogue: 0,0:30:52.22,0:30:55.83,Default,,0000,0000,0000,,Also, if we pick X equal to 1,\Nthe last term vanish is because
Dialogue: 0,0:30:55.83,0:30:59.96,Default,,0000,0000,0000,,we have a 1 - 1 which is zero\Nand will be just left with the
Dialogue: 0,0:30:59.96,0:31:03.93,Default,,0000,0000,0000,,term involving be. So by letting\NXP, one will have one.
Dialogue: 0,0:31:04.61,0:31:07.63,Default,,0000,0000,0000,,On the left is equal to 0.
Dialogue: 0,0:31:08.20,0:31:12.07,Default,,0000,0000,0000,,One and one here is 22B.
Dialogue: 0,0:31:12.07,0:31:19.09,Default,,0000,0000,0000,,And the last term vanishes. In\Nother words, B is equal to 1/2.
Dialogue: 0,0:31:20.38,0:31:24.98,Default,,0000,0000,0000,,What's another sensible value to\Npick for X? Well, if we let X
Dialogue: 0,0:31:24.98,0:31:26.04,Default,,0000,0000,0000,,equal minus one.
Dialogue: 0,0:31:26.78,0:31:30.64,Default,,0000,0000,0000,,X being minus one will mean that\Nthis term vanish is minus one
Dialogue: 0,0:31:30.64,0:31:31.83,Default,,0000,0000,0000,,plus One is 0.
Dialogue: 0,0:31:32.52,0:31:36.95,Default,,0000,0000,0000,,This term will vanish minus one,\Nplus one is zero and will be
Dialogue: 0,0:31:36.95,0:31:41.39,Default,,0000,0000,0000,,able to find see. So I'm going\Nto let X be minus one.
Dialogue: 0,0:31:41.62,0:31:44.71,Default,,0000,0000,0000,,Will still have the one on the
Dialogue: 0,0:31:44.71,0:31:47.89,Default,,0000,0000,0000,,left. When X is minus one, this
Dialogue: 0,0:31:47.89,0:31:50.11,Default,,0000,0000,0000,,goes. This goes.
Dialogue: 0,0:31:50.68,0:31:55.38,Default,,0000,0000,0000,,And on the right hand side this\Nterm here will have minus 1 - 1
Dialogue: 0,0:31:55.38,0:31:59.76,Default,,0000,0000,0000,,is minus two we square it will\Nget plus four, so will get plus
Dialogue: 0,0:31:59.76,0:32:04.100,Default,,0000,0000,0000,,4C. In other words, see is going\Nto be 1/4.
Dialogue: 0,0:32:06.07,0:32:11.11,Default,,0000,0000,0000,,So we've got be. We've got C and\Nnow we need to find a value for
Dialogue: 0,0:32:11.11,0:32:15.20,Default,,0000,0000,0000,,a. Now we can substitute any\Nother value we like in here, so
Dialogue: 0,0:32:15.20,0:32:18.98,Default,,0000,0000,0000,,I'm actually going to pick X\Nequals 0. It's a nice simple
Dialogue: 0,0:32:18.98,0:32:23.08,Default,,0000,0000,0000,,value. Effects is zero, will\Nhave one on the left if X is
Dialogue: 0,0:32:23.08,0:32:26.23,Default,,0000,0000,0000,,zero there and there will be\Nleft with minus one.
Dialogue: 0,0:32:26.79,0:32:32.40,Default,,0000,0000,0000,,Minus 1 * 1 is minus 1 - 1\Ntimes a is minus a.
Dialogue: 0,0:32:33.10,0:32:38.84,Default,,0000,0000,0000,,Thanks is 0 here. Will just left\Nwith B Times one which is be.
Dialogue: 0,0:32:38.85,0:32:44.42,Default,,0000,0000,0000,,And if X is 0 here will have\Nminus one squared, which is plus
Dialogue: 0,0:32:44.42,0:32:50.39,Default,,0000,0000,0000,,one plus One Times C Plus C. Now\Nwe already know values for B and
Dialogue: 0,0:32:50.39,0:32:55.96,Default,,0000,0000,0000,,for C, so we substitute these in\Nwill have one is equal to minus
Dialogue: 0,0:32:55.96,0:33:00.34,Default,,0000,0000,0000,,A plus B which is 1/2 plus C\Nwhich is 1/4.
Dialogue: 0,0:33:01.49,0:33:06.15,Default,,0000,0000,0000,,So rearranging this will have\Nthat a is equal to.
Dialogue: 0,0:33:06.73,0:33:12.26,Default,,0000,0000,0000,,Well, with a half and a quarter,\Nthat's 3/4 and the one from this
Dialogue: 0,0:33:12.26,0:33:17.79,Default,,0000,0000,0000,,side over the other side is\Nminus one. 3/4 - 1 is going to
Dialogue: 0,0:33:17.79,0:33:18.98,Default,,0000,0000,0000,,be minus 1/4.
Dialogue: 0,0:33:18.98,0:33:25.97,Default,,0000,0000,0000,,So there we have our values for\Na for B and for C, so a being
Dialogue: 0,0:33:25.97,0:33:29.03,Default,,0000,0000,0000,,minus 1/4 will go back in there.
Dialogue: 0,0:33:29.33,0:33:34.06,Default,,0000,0000,0000,,Be being a half will go back in\Nhere and see being a quarter
Dialogue: 0,0:33:34.06,0:33:38.12,Default,,0000,0000,0000,,will go back in there and then\Nwe'll have three separate pieces
Dialogue: 0,0:33:38.12,0:33:41.50,Default,,0000,0000,0000,,of integration to do in order to\Ncomplete the problem.
Dialogue: 0,0:33:42.46,0:33:46.47,Default,,0000,0000,0000,,Let me write these all down\Nagain. I'm going to write the
Dialogue: 0,0:33:46.47,0:33:49.47,Default,,0000,0000,0000,,integral out and write these\Nthree separate ones down.
Dialogue: 0,0:33:49.49,0:33:56.83,Default,,0000,0000,0000,,So we had that the integral of\None over X minus one squared X
Dialogue: 0,0:33:56.83,0:34:03.11,Default,,0000,0000,0000,,Plus one, all integrated with\Nrespect to X, is going to equal.
Dialogue: 0,0:34:03.86,0:34:09.94,Default,,0000,0000,0000,,Minus 1/4 the integral of one\Nover X minus one.
Dialogue: 0,0:34:10.61,0:34:17.92,Default,,0000,0000,0000,,They'll be plus 1/2 integral of\None over X minus 1 squared.
Dialogue: 0,0:34:18.48,0:34:24.38,Default,,0000,0000,0000,,And they'll\Nbe 1/4
Dialogue: 0,0:34:24.38,0:34:27.34,Default,,0000,0000,0000,,over X
Dialogue: 0,0:34:27.34,0:34:33.42,Default,,0000,0000,0000,,plus one.\NAnd we want
Dialogue: 0,0:34:33.42,0:34:36.50,Default,,0000,0000,0000,,to DX in
Dialogue: 0,0:34:36.50,0:34:42.03,Default,,0000,0000,0000,,every term. So we've\Nused partial fractions to split
Dialogue: 0,0:34:42.03,0:34:45.82,Default,,0000,0000,0000,,this integrand into three\Nseparate terms, and will try and
Dialogue: 0,0:34:45.82,0:34:49.61,Default,,0000,0000,0000,,finish this off. Now. The first\Nintegral straightforward when we
Dialogue: 0,0:34:49.61,0:34:54.16,Default,,0000,0000,0000,,integrate one over X minus one\Nwill end up with just the
Dialogue: 0,0:34:54.16,0:34:57.57,Default,,0000,0000,0000,,natural logarithm of the\Ndenominator, so we'll end up
Dialogue: 0,0:34:57.57,0:35:00.98,Default,,0000,0000,0000,,with minus 1/4 natural\Nlogarithm. The modulus of X
Dialogue: 0,0:35:00.98,0:35:06.21,Default,,0000,0000,0000,,minus one. To integrate this\Nterm, we're integrating one over
Dialogue: 0,0:35:06.21,0:35:10.62,Default,,0000,0000,0000,,X minus one squared. Let me just\Ndo that separately.
Dialogue: 0,0:35:10.77,0:35:16.36,Default,,0000,0000,0000,,To integrate one over X minus\None all squared, we make a
Dialogue: 0,0:35:16.36,0:35:20.09,Default,,0000,0000,0000,,substitution and let you equals\NX minus one.
Dialogue: 0,0:35:20.76,0:35:25.28,Default,,0000,0000,0000,,Do you then will equal du DX,\Nwhich is just one times DX, so
Dialogue: 0,0:35:25.28,0:35:27.87,Default,,0000,0000,0000,,do you will be the same as DX?
Dialogue: 0,0:35:28.45,0:35:32.24,Default,,0000,0000,0000,,The integral will become the\Nintegral of one over.
Dialogue: 0,0:35:32.98,0:35:35.68,Default,,0000,0000,0000,,X minus one squared\Nwill be you squared.
Dialogue: 0,0:35:36.82,0:35:38.44,Default,,0000,0000,0000,,And DX is D.
Dialogue: 0,0:35:39.21,0:35:42.67,Default,,0000,0000,0000,,Now, this is straightforward to\Nfinish because this is
Dialogue: 0,0:35:42.67,0:35:44.97,Default,,0000,0000,0000,,integrating you to the minus 2.
Dialogue: 0,0:35:45.63,0:35:49.04,Default,,0000,0000,0000,,Increase the power by one\Nbecomes you to the minus 1
Dialogue: 0,0:35:49.04,0:35:50.59,Default,,0000,0000,0000,,divided by the new power.
Dialogue: 0,0:35:51.09,0:35:52.81,Default,,0000,0000,0000,,And add a constant of\Nintegration.
Dialogue: 0,0:35:54.01,0:35:59.20,Default,,0000,0000,0000,,So when we do this, integration\Nwill end up with minus one over
Dialogue: 0,0:35:59.20,0:36:01.48,Default,,0000,0000,0000,,you. Which is minus one over.
Dialogue: 0,0:36:02.25,0:36:04.01,Default,,0000,0000,0000,,X minus one.
Dialogue: 0,0:36:04.66,0:36:10.86,Default,,0000,0000,0000,,And we can put that back into\Nhere now. So the integral half
Dialogue: 0,0:36:10.86,0:36:16.11,Default,,0000,0000,0000,,integral of one over X minus one\Nsquared will be 1/2.
Dialogue: 0,0:36:16.21,0:36:19.88,Default,,0000,0000,0000,,All that we've got down here,\Nwhich is minus one over X minus
Dialogue: 0,0:36:19.88,0:36:24.96,Default,,0000,0000,0000,,one. The constant of integration\Nwill add another very end. This
Dialogue: 0,0:36:24.96,0:36:27.32,Default,,0000,0000,0000,,integral is going to be 1/4.
Dialogue: 0,0:36:27.89,0:36:31.89,Default,,0000,0000,0000,,The natural logarithm of the\Nmodulus of X Plus One and then a
Dialogue: 0,0:36:31.89,0:36:33.74,Default,,0000,0000,0000,,single constant for all of that.
Dialogue: 0,0:36:35.11,0:36:39.49,Default,,0000,0000,0000,,Let me just tidy up this little\Na little bit. The two logarithm
Dialogue: 0,0:36:39.49,0:36:43.20,Default,,0000,0000,0000,,terms can be combined. We've got\Na quarter of this logarithm.
Dialogue: 0,0:36:43.20,0:36:46.90,Default,,0000,0000,0000,,Subtract 1/4 of this logarithm,\Nso together we've got a quarter
Dialogue: 0,0:36:46.90,0:36:50.61,Default,,0000,0000,0000,,the logarithm. If we use the\Nlaws of logarithms, we've gotta
Dialogue: 0,0:36:50.61,0:36:54.66,Default,,0000,0000,0000,,log subtracted log. So we want\Nthe first term X Plus one
Dialogue: 0,0:36:54.66,0:36:56.68,Default,,0000,0000,0000,,divided by the second term X
Dialogue: 0,0:36:56.68,0:37:03.28,Default,,0000,0000,0000,,minus one. And then this term\Ncan be written as minus 1/2.
Dialogue: 0,0:37:03.28,0:37:06.06,Default,,0000,0000,0000,,One over X minus one.
Dialogue: 0,0:37:06.06,0:37:09.95,Default,,0000,0000,0000,,Is a constant of integration\Nat the end, and that's the
Dialogue: 0,0:37:09.95,0:37:10.66,Default,,0000,0000,0000,,integration complete.
Dialogue: 0,0:37:11.96,0:37:18.44,Default,,0000,0000,0000,,Now let's look at an example in\Nwhich we have an improper
Dialogue: 0,0:37:18.44,0:37:21.27,Default,,0000,0000,0000,,fraction. Suppose we have this
Dialogue: 0,0:37:21.27,0:37:27.47,Default,,0000,0000,0000,,integral. The\Ndegree of
Dialogue: 0,0:37:27.47,0:37:30.73,Default,,0000,0000,0000,,the numerator
Dialogue: 0,0:37:30.73,0:37:36.33,Default,,0000,0000,0000,,is 3.\NThe degree of the
Dialogue: 0,0:37:36.33,0:37:37.56,Default,,0000,0000,0000,,denominator is 2.
Dialogue: 0,0:37:38.61,0:37:41.62,Default,,0000,0000,0000,,3 being greater than two\Nmeans that this is an
Dialogue: 0,0:37:41.62,0:37:42.22,Default,,0000,0000,0000,,improper fraction.
Dialogue: 0,0:37:43.33,0:37:46.39,Default,,0000,0000,0000,,Improper fractions requires\Nspecial treatment and the first
Dialogue: 0,0:37:46.39,0:37:49.82,Default,,0000,0000,0000,,thing we do is division\Npolynomial division. Now, if
Dialogue: 0,0:37:49.82,0:37:53.64,Default,,0000,0000,0000,,you're not happy with Long\NDivision of polynomials, then I
Dialogue: 0,0:37:53.64,0:37:57.46,Default,,0000,0000,0000,,would suggest that you look\Nagain at the video called
Dialogue: 0,0:37:57.46,0:38:00.87,Default,,0000,0000,0000,,polynomial division. When\Nexamples like this are done very
Dialogue: 0,0:38:00.87,0:38:05.12,Default,,0000,0000,0000,,thoroughly, but what we want to\Ndo is see how many times X minus
Dialogue: 0,0:38:05.12,0:38:06.95,Default,,0000,0000,0000,,X squared minus four will divide
Dialogue: 0,0:38:06.95,0:38:13.47,Default,,0000,0000,0000,,into X cubed. The way we do\Nthis polynomial division is we
Dialogue: 0,0:38:13.47,0:38:19.63,Default,,0000,0000,0000,,say how many times does X\Nsquared divided into X cubed.
Dialogue: 0,0:38:20.33,0:38:24.88,Default,,0000,0000,0000,,That's like asking How many\Ntimes X squared will go into X
Dialogue: 0,0:38:24.88,0:38:29.05,Default,,0000,0000,0000,,cubed, and clearly when X\Nsquared is divided into X cubed,
Dialogue: 0,0:38:29.05,0:38:30.94,Default,,0000,0000,0000,,the answer is just X.
Dialogue: 0,0:38:31.92,0:38:35.65,Default,,0000,0000,0000,,So X squared goes into X\Ncubed X times and we write
Dialogue: 0,0:38:35.65,0:38:36.90,Default,,0000,0000,0000,,the solution down there.
Dialogue: 0,0:38:38.13,0:38:42.19,Default,,0000,0000,0000,,We take what we have just\Nwritten down and multiply it by
Dialogue: 0,0:38:42.19,0:38:45.57,Default,,0000,0000,0000,,everything here. So X times X\Nsquared is X cubed.
Dialogue: 0,0:38:46.13,0:38:49.59,Default,,0000,0000,0000,,X times minus four is minus 4X.
Dialogue: 0,0:38:50.37,0:38:51.88,Default,,0000,0000,0000,,And then we subtract.
Dialogue: 0,0:38:52.69,0:38:55.47,Default,,0000,0000,0000,,X cubed minus X cubed vanish is.
Dialogue: 0,0:38:56.04,0:39:00.38,Default,,0000,0000,0000,,With no access here, and we're\Nsubtracting minus 4X, which is
Dialogue: 0,0:39:00.38,0:39:01.57,Default,,0000,0000,0000,,like adding 4X.
Dialogue: 0,0:39:03.11,0:39:08.14,Default,,0000,0000,0000,,This means that when we divide X\Nsquared minus four into X cubed,
Dialogue: 0,0:39:08.14,0:39:13.56,Default,,0000,0000,0000,,we get a whole part X and a\Nremainder 4X. In other words, X
Dialogue: 0,0:39:13.56,0:39:18.20,Default,,0000,0000,0000,,cubed divided by X squared minus\Nfour can be written as X.
Dialogue: 0,0:39:19.36,0:39:25.10,Default,,0000,0000,0000,,Plus 4X divided by X\Nsquared minus 4.
Dialogue: 0,0:39:25.27,0:39:30.24,Default,,0000,0000,0000,,So in order to tackle this\Nintegration, we've done the long
Dialogue: 0,0:39:30.24,0:39:34.76,Default,,0000,0000,0000,,division and we're left with two\Nseparate integrals to workout.
Dialogue: 0,0:39:35.64,0:39:37.98,Default,,0000,0000,0000,,Now this one is going to be\Nstraightforward. Clearly you're
Dialogue: 0,0:39:37.98,0:39:41.02,Default,,0000,0000,0000,,just integrating X, it's going\Nto be X squared over 2. This is
Dialogue: 0,0:39:41.02,0:39:43.60,Default,,0000,0000,0000,,a bit more problematic. Let's\Nhave a look at this again.
Dialogue: 0,0:39:44.42,0:39:48.08,Default,,0000,0000,0000,,This is now a proper fraction\Nwhere the degree of the
Dialogue: 0,0:39:48.08,0:39:51.41,Default,,0000,0000,0000,,numerator is one with the next\Nto the one here.
Dialogue: 0,0:39:52.85,0:39:54.51,Default,,0000,0000,0000,,The degree of the denominator is
Dialogue: 0,0:39:54.51,0:39:56.81,Default,,0000,0000,0000,,2. So it's a proper fraction.
Dialogue: 0,0:39:58.13,0:40:00.99,Default,,0000,0000,0000,,It looks as though we've got a\Nquadratic factor in the
Dialogue: 0,0:40:00.99,0:40:05.69,Default,,0000,0000,0000,,denominator. But in fact, X\Nsquared minus four will
Dialogue: 0,0:40:05.69,0:40:10.42,Default,,0000,0000,0000,,factorize. It's in fact, the\Ndifference of two squares. So we
Dialogue: 0,0:40:10.42,0:40:16.44,Default,,0000,0000,0000,,can write 4X over X squared\Nminus four as 4X over X minus 2X
Dialogue: 0,0:40:16.44,0:40:20.74,Default,,0000,0000,0000,,plus two. So the quadratic will\Nactually factorize and you'll
Dialogue: 0,0:40:20.74,0:40:23.32,Default,,0000,0000,0000,,see. Now we've got two linear
Dialogue: 0,0:40:23.32,0:40:28.48,Default,,0000,0000,0000,,factors. We can express this in\Npartial fractions. Let's do
Dialogue: 0,0:40:28.48,0:40:35.02,Default,,0000,0000,0000,,that. We've got 4X over X, minus\Ntwo X +2, and because each of
Dialogue: 0,0:40:35.02,0:40:40.16,Default,,0000,0000,0000,,these are linear factors, the\Nappropriate form is going to be
Dialogue: 0,0:40:40.16,0:40:42.49,Default,,0000,0000,0000,,a over X minus 2.
Dialogue: 0,0:40:43.16,0:40:46.98,Default,,0000,0000,0000,,Plus B over X +2.
Dialogue: 0,0:40:49.52,0:40:54.99,Default,,0000,0000,0000,,We add these together using a\Ncommon dumped common denominator
Dialogue: 0,0:40:54.99,0:41:02.65,Default,,0000,0000,0000,,X minus two X +2 and will\Nget a X +2 plus BX minus
Dialogue: 0,0:41:02.65,0:41:09.21,Default,,0000,0000,0000,,2. All over the common\Ndenominator, X minus two X +2.
Dialogue: 0,0:41:12.31,0:41:18.99,Default,,0000,0000,0000,,Now the left hand side and the\Nright hand side of the same. The
Dialogue: 0,0:41:18.99,0:41:23.76,Default,,0000,0000,0000,,denominators are already the\Nsame, so the numerators must be
Dialogue: 0,0:41:23.76,0:41:27.57,Default,,0000,0000,0000,,the same. 4X must equal a X +2
Dialogue: 0,0:41:27.57,0:41:29.79,Default,,0000,0000,0000,,plus B. 6 - 2.
Dialogue: 0,0:41:30.85,0:41:35.71,Default,,0000,0000,0000,,This is the equation that we can\Nuse to find the values for A&B.
Dialogue: 0,0:41:37.55,0:41:39.46,Default,,0000,0000,0000,,Let me write that down again.
Dialogue: 0,0:41:39.99,0:41:45.76,Default,,0000,0000,0000,,We've got 4X is\Na X +2.
Dialogue: 0,0:41:46.44,0:41:51.02,Default,,0000,0000,0000,,Plus BX minus\N2.
Dialogue: 0,0:41:52.63,0:41:56.84,Default,,0000,0000,0000,,What's a sensible value to put\Nin for X? Well, if we choose, X
Dialogue: 0,0:41:56.84,0:41:58.95,Default,,0000,0000,0000,,is 2, will lose the second term.
Dialogue: 0,0:41:59.62,0:42:02.73,Default,,0000,0000,0000,,So X is 2 is a good\Nvalue to put in here.
Dialogue: 0,0:42:03.94,0:42:09.10,Default,,0000,0000,0000,,Faxes 2 on the left hand side\Nwill have 4 * 2, which is 8.
Dialogue: 0,0:42:10.07,0:42:15.41,Default,,0000,0000,0000,,If X is 2, will have 2 +\N2, which is 44A.
Dialogue: 0,0:42:15.41,0:42:17.94,Default,,0000,0000,0000,,And this term will vanish.
Dialogue: 0,0:42:19.00,0:42:22.20,Default,,0000,0000,0000,,So 8 equals 4A.
Dialogue: 0,0:42:22.38,0:42:27.34,Default,,0000,0000,0000,,A will equal 2 and that's the\Nvalue for A.
Dialogue: 0,0:42:29.20,0:42:32.68,Default,,0000,0000,0000,,What's another sensible value to\Nput in for X? Well, if X is
Dialogue: 0,0:42:32.68,0:42:36.44,Default,,0000,0000,0000,,minus two, will have minus 2 +\N2, which is zero. Will lose this
Dialogue: 0,0:42:36.44,0:42:44.20,Default,,0000,0000,0000,,term. So if X is\Nminus two, will have minus 8
Dialogue: 0,0:42:44.20,0:42:46.18,Default,,0000,0000,0000,,on the left.
Dialogue: 0,0:42:46.19,0:42:47.71,Default,,0000,0000,0000,,This first term will vanish.
Dialogue: 0,0:42:48.49,0:42:54.49,Default,,0000,0000,0000,,An ex being minus two here will\Nhave minus 2 - 2 is minus four
Dialogue: 0,0:42:54.49,0:42:55.69,Default,,0000,0000,0000,,so minus 4B.
Dialogue: 0,0:42:55.69,0:43:01.11,Default,,0000,0000,0000,,So be must also be\Nequal to two.
Dialogue: 0,0:43:01.12,0:43:06.16,Default,,0000,0000,0000,,So now we've got values for A\Nand for be both equal to two.
Dialogue: 0,0:43:06.16,0:43:08.68,Default,,0000,0000,0000,,Let's take us back and see what
Dialogue: 0,0:43:08.68,0:43:11.11,Default,,0000,0000,0000,,that means. It means that when
Dialogue: 0,0:43:11.11,0:43:14.74,Default,,0000,0000,0000,,we express. This quantity\N4X over X squared minus
Dialogue: 0,0:43:14.74,0:43:17.87,Default,,0000,0000,0000,,four in partial fractions\Nlike this, the values of
Dialogue: 0,0:43:17.87,0:43:19.96,Default,,0000,0000,0000,,A&B are both equal to two.
Dialogue: 0,0:43:21.14,0:43:26.74,Default,,0000,0000,0000,,So the integral were working out\Nis the integral of X +2 over X
Dialogue: 0,0:43:26.74,0:43:29.54,Default,,0000,0000,0000,,minus 2 + 2 over X +2.
Dialogue: 0,0:43:29.55,0:43:35.69,Default,,0000,0000,0000,,So there we\Nare. We've now
Dialogue: 0,0:43:35.69,0:43:41.84,Default,,0000,0000,0000,,got three separate\Nintegrals to evaluate,
Dialogue: 0,0:43:41.84,0:43:47.98,Default,,0000,0000,0000,,and it's straightforward\Nto finish this
Dialogue: 0,0:43:47.98,0:43:54.54,Default,,0000,0000,0000,,off. The integral of X\Nis X squared divided by two.
Dialogue: 0,0:43:55.65,0:44:02.48,Default,,0000,0000,0000,,The integral of two over X minus\N2 equals 2 natural logarithm of
Dialogue: 0,0:44:02.48,0:44:05.62,Default,,0000,0000,0000,,the modulus of X minus 2.
Dialogue: 0,0:44:06.43,0:44:13.93,Default,,0000,0000,0000,,And the integral of two over X\N+2 is 2 natural logarithm of X
Dialogue: 0,0:44:13.93,0:44:16.61,Default,,0000,0000,0000,,+2 plus a constant of
Dialogue: 0,0:44:16.61,0:44:20.95,Default,,0000,0000,0000,,integration. If we wanted to do,\Nwe could use the laws of
Dialogue: 0,0:44:20.95,0:44:23.62,Default,,0000,0000,0000,,logarithms to combine these log\Nrhythmic terms here, but I'll
Dialogue: 0,0:44:23.62,0:44:24.95,Default,,0000,0000,0000,,leave the answer like that.