1 00:00:00,900 --> 00:00:04,540 In this video, we're going to have a look at how we can 2 00:00:04,540 --> 00:00:06,721 integrate algebraic fractions. The sorts of 3 00:00:06,721 --> 00:00:09,214 fractions that we're going to integrate and like these 4 00:00:09,214 --> 00:00:09,491 here. 5 00:00:10,590 --> 00:00:13,990 Now, superficially, they will look very similar, but there are 6 00:00:13,990 --> 00:00:17,730 important differences which I'd like to point out when you come 7 00:00:17,730 --> 00:00:20,790 to tackle a problem of integrating a fraction like 8 00:00:20,790 --> 00:00:24,190 this, it's important that you can look for certain features, 9 00:00:24,190 --> 00:00:25,890 for example, in this first 10 00:00:25,890 --> 00:00:30,610 example. In the denominator we have what we call 2 linear 11 00:00:30,610 --> 00:00:33,976 factors. These two linear factors are two different linear 12 00:00:33,976 --> 00:00:38,090 factors. When I say linear factors, I mean there's no X 13 00:00:38,090 --> 00:00:42,578 squared, no ex cubes, nothing like that in it. These are just 14 00:00:42,578 --> 00:00:47,814 of the form a X Plus B linear factors. A constant times X plus 15 00:00:47,814 --> 00:00:50,806 another constant. So with two linear factors here. 16 00:00:51,870 --> 00:00:55,965 This example is also got linear factors in clearly X Plus One is 17 00:00:55,965 --> 00:01:00,720 a linear factor. X minus one is a linear factor, but the fact 18 00:01:00,720 --> 00:01:04,440 that I've got an X minus one squared means that we've really 19 00:01:04,440 --> 00:01:08,470 got X minus one times X minus 1 two linear factors within here. 20 00:01:08,470 --> 00:01:11,880 So we call this an example of a repeated linear factor. 21 00:01:13,150 --> 00:01:18,600 Linear factors, repeated linear factors. Over here we've got a 22 00:01:18,600 --> 00:01:23,505 quadratic. Now this particular quadratic will not factorize and 23 00:01:23,505 --> 00:01:28,955 because it won't factorize, we call it an irreducible quadratic 24 00:01:28,955 --> 00:01:33,700 factor. This final example here has also got a quadratic factor 25 00:01:33,700 --> 00:01:37,330 in the denominator, but unlike the previous one, this one will 26 00:01:37,330 --> 00:01:40,630 in fact factorize because X squared minus four is actually 27 00:01:40,630 --> 00:01:44,920 the difference of two squares and we can write this as X minus 28 00:01:44,920 --> 00:01:48,880 2X plus two. So whilst it might have originally looked like a 29 00:01:48,880 --> 00:01:52,510 quadratic factor, it was in fact to linear factors, so that's 30 00:01:52,510 --> 00:01:56,140 what those are one of the important things we should be 31 00:01:56,140 --> 00:01:59,440 looking for when we come to integrate quantities like these. 32 00:01:59,440 --> 00:02:00,760 Weather we've got linear 33 00:02:00,760 --> 00:02:03,535 factors. Repeated linear factors, irreducible quadratic 34 00:02:03,535 --> 00:02:06,902 factors, or quadratic factors that will factorize. 35 00:02:07,440 --> 00:02:10,560 Something else which is important as well, is to examine 36 00:02:10,560 --> 00:02:14,304 the degree of the numerator and the degree of the denominator in 37 00:02:14,304 --> 00:02:17,112 each of these fractions. Remember, the degree is the 38 00:02:17,112 --> 00:02:20,544 highest power, so for example in the denominator of this example 39 00:02:20,544 --> 00:02:24,288 here, if we multiplied it all out, we actually get the highest 40 00:02:24,288 --> 00:02:27,720 power as three, because when we multiply the first terms out 41 00:02:27,720 --> 00:02:31,776 will get an X squared, and when we multiply it with the SEC 42 00:02:31,776 --> 00:02:36,144 bracket, X Plus, one will end up with an X cubed. So the degree 43 00:02:36,144 --> 00:02:38,016 of the denominator there is 3. 44 00:02:38,670 --> 00:02:42,954 The degree of the numerator is 0 because we can think of this as 45 00:02:42,954 --> 00:02:44,484 One X to the 0. 46 00:02:46,040 --> 00:02:50,345 In the first case, we've got an X to the one here, so the degree 47 00:02:50,345 --> 00:02:53,789 of the numerator there is one, and if we multiply the brackets 48 00:02:53,789 --> 00:02:57,233 out, the degree of the of the denominator will be two. Will 49 00:02:57,233 --> 00:03:00,964 get a quadratic term in here. In this case, the degree of the 50 00:03:00,964 --> 00:03:05,262 numerator is 0. And in this case, the degree of the 51 00:03:05,262 --> 00:03:09,630 denominator is to the highest power is too. So in all of 52 00:03:09,630 --> 00:03:13,634 these cases, the degree of the numerator is less than the 53 00:03:13,634 --> 00:03:17,274 degree of the denominator, and we call fractions like these 54 00:03:17,274 --> 00:03:18,002 proper fractions. 55 00:03:20,110 --> 00:03:24,114 On the other hand, if we look at this final example, the degree 56 00:03:24,114 --> 00:03:27,810 of the numerator is 3, whereas the degree of the denominator is 57 00:03:27,810 --> 00:03:31,814 too. So in this case, the degree of the numerator is greater than 58 00:03:31,814 --> 00:03:35,202 the degree of the denominator, and this is what's called an 59 00:03:35,202 --> 00:03:38,670 improper fraction. Now when we start to integrate quantities 60 00:03:38,670 --> 00:03:41,970 like this will need to examine whether we're dealing with 61 00:03:41,970 --> 00:03:45,270 proper fractions or improper fractions, and then, as I said 62 00:03:45,270 --> 00:03:49,230 before, will need to look at all the factors in the denominator. 63 00:03:49,800 --> 00:03:53,485 Will also need to call appan techniques in the theory of 64 00:03:53,485 --> 00:03:56,835 partial fractions. There is a video on partial fractions and 65 00:03:56,835 --> 00:03:59,515 you may wish to refer to that if 66 00:03:59,515 --> 00:04:04,090 necessary. If you have a linear factor in the denominator, this 67 00:04:04,090 --> 00:04:08,254 will lead to a partial fraction of this form. A constant over 68 00:04:08,254 --> 00:04:09,295 the linear factor. 69 00:04:10,100 --> 00:04:13,740 If you have a repeated linear factor in the denominator, 70 00:04:13,740 --> 00:04:16,652 you'll need two partial fractions. A constant over 71 00:04:16,652 --> 00:04:19,928 the factor and a constant over the factor squared. 72 00:04:21,840 --> 00:04:25,460 Finally, if you have a quadratic factor which is irreducible, 73 00:04:25,460 --> 00:04:29,804 you'll need to write a partial fraction of the form a constant 74 00:04:29,804 --> 00:04:33,062 times X plus another constant over the irreducible quadratic 75 00:04:33,062 --> 00:04:36,682 factor, so will certainly be calling upon the techniques of 76 00:04:36,682 --> 00:04:41,616 partial fractions. Will also need to call appan. Lots of 77 00:04:41,616 --> 00:04:42,987 techniques and integration. 78 00:04:42,990 --> 00:04:46,554 I'm just going to mention just two or three here, which will 79 00:04:46,554 --> 00:04:49,821 need to use as we proceed through the examples of 1st. 80 00:04:49,821 --> 00:04:53,088 Crucial result is the standard result, which says that if you 81 00:04:53,088 --> 00:04:56,058 have an integral consisting of a function in the denominator. 82 00:04:56,770 --> 00:05:00,810 And it's derivative in the numerator. Then the result is 83 00:05:00,810 --> 00:05:05,254 the logarithm of the modulus of the function in the denominator. 84 00:05:05,254 --> 00:05:10,910 So for example, if I ask you to integrate one over X plus one 85 00:05:10,910 --> 00:05:12,526 with respect to X. 86 00:05:13,320 --> 00:05:16,818 Then clearly the function in the denominator is X plus one. 87 00:05:17,390 --> 00:05:20,668 And its derivative is one which appears in the numerator. So 88 00:05:20,668 --> 00:05:22,456 we've an example of this form. 89 00:05:23,640 --> 00:05:27,372 So the resulting integral is the logarithm of the modulus of the 90 00:05:27,372 --> 00:05:30,482 function that was in the denominator, which is X plus 91 00:05:30,482 --> 00:05:34,321 one. Plus a constant of integration, so we will need 92 00:05:34,321 --> 00:05:37,830 that result very frequently in the examples which are going to 93 00:05:37,830 --> 00:05:41,339 follow will also need some standard results and one of the 94 00:05:41,339 --> 00:05:44,848 standard results I will call appan is this one. The integral 95 00:05:44,848 --> 00:05:48,995 of one over a squared plus X squared is one over a inverse 96 00:05:48,995 --> 00:05:51,228 tan of X over a plus C. 97 00:05:52,400 --> 00:05:55,821 Results like this can be found in tables of standard integrals. 98 00:05:57,120 --> 00:06:00,850 Finally, we need to integrate quantities like this and you'll 99 00:06:00,850 --> 00:06:04,580 need to do this probably using integration by substitution. An 100 00:06:04,580 --> 00:06:08,683 integral like this can be worked out by making the substitution 101 00:06:08,683 --> 00:06:10,548 you equals X minus one. 102 00:06:11,260 --> 00:06:16,604 So that the differential du is du DX. 103 00:06:16,610 --> 00:06:22,790 DX, which in this case is du DX, will be just one. 104 00:06:22,850 --> 00:06:27,470 So do you is DX and that's integral. Then will become 105 00:06:27,470 --> 00:06:30,830 the integral of one over you, squared du. 106 00:06:32,080 --> 00:06:36,588 One over you squared is the same as the integral of you to the 107 00:06:36,588 --> 00:06:40,130 minus 2D U, which you can solve by integrating increasing the 108 00:06:40,130 --> 00:06:44,638 power BI want to give you you to the minus one over minus one. 109 00:06:45,290 --> 00:06:46,770 Plus a constant of integration. 110 00:06:48,180 --> 00:06:51,912 This can be finished off by changing the you back to the 111 00:06:51,912 --> 00:06:55,644 original variable X minus one and that will give us X minus 112 00:06:55,644 --> 00:06:57,821 one to the minus one over minus 113 00:06:57,821 --> 00:07:02,868 one plus C. Which is the same as minus one over X minus one plus 114 00:07:02,868 --> 00:07:04,891 C, which is the results I have 115 00:07:04,891 --> 00:07:09,384 here. So what I'm saying is that throughout the rest of this unit 116 00:07:09,384 --> 00:07:12,708 will need to call Appan lots of different techniques to be able 117 00:07:12,708 --> 00:07:13,816 to perform the integrals. 118 00:07:14,540 --> 00:07:16,020 As we shall see. 119 00:07:16,590 --> 00:07:19,902 Let's look at the 120 00:07:19,902 --> 00:07:26,492 first example. Suppose we want to integrate this algebraic 121 00:07:26,492 --> 00:07:29,772 fraction. 6 / 2 122 00:07:29,772 --> 00:07:33,245 minus X. X 123 00:07:33,245 --> 00:07:38,140 +3 DX 124 00:07:39,300 --> 00:07:42,990 The first thing we do is we look at the object we've got and try 125 00:07:42,990 --> 00:07:45,696 to ask ourselves, are we dealing with a proper or improper 126 00:07:45,696 --> 00:07:48,156 fraction and what are the factors in the denominator like? 127 00:07:49,200 --> 00:07:53,208 Well, if we multiply the power, the brackets at the bottom will 128 00:07:53,208 --> 00:07:57,884 find that the highest power of X is X squared, so the degree of 129 00:07:57,884 --> 00:07:59,220 the denominator is 2. 130 00:08:00,430 --> 00:08:04,318 The highest power in the numerator is one. This is an X 131 00:08:04,318 --> 00:08:08,530 to the power one, so the degree of the numerator is one because 132 00:08:08,530 --> 00:08:12,418 the degree of the numerator is less than the degree of the 133 00:08:12,418 --> 00:08:16,460 denominator. This is an example of a proper fraction. 134 00:08:17,040 --> 00:08:22,745 Both of these factors in the denominator. 135 00:08:23,380 --> 00:08:24,658 Are linear factors. 136 00:08:25,250 --> 00:08:31,800 So we're dealing with a proper fraction with linear factors. 137 00:08:31,820 --> 00:08:37,787 The way we proceed is to take this fraction and express it in 138 00:08:37,787 --> 00:08:41,918 partial fractions. So I'll start with the fraction again. 139 00:08:42,770 --> 00:08:49,043 And express it in the appropriate form of partial 140 00:08:49,043 --> 00:08:51,831 fractions. Now because it's 141 00:08:51,831 --> 00:08:56,206 proper. And because we've got linear factors, the appropriate 142 00:08:56,206 --> 00:09:00,573 form is to have a constant over the first linear factor. 143 00:09:01,560 --> 00:09:06,376 Plus another constant over the second linear factor. 144 00:09:07,580 --> 00:09:11,529 Our task now is to find values for the constants A&B. 145 00:09:12,220 --> 00:09:15,980 Once we've done that, will be able to evaluate this 146 00:09:15,980 --> 00:09:18,236 integral by evaluating these two separately. 147 00:09:19,480 --> 00:09:24,145 So to find A and be the first thing we do is we add these 148 00:09:24,145 --> 00:09:25,389 two fractions together again. 149 00:09:26,440 --> 00:09:31,276 Remember that to add 2 fractions together, we've got to give them 150 00:09:31,276 --> 00:09:34,903 the same denominator. They've got to have a common 151 00:09:34,903 --> 00:09:38,530 denominator. The common denominator is going to be made 152 00:09:38,530 --> 00:09:41,351 up of the two factors. 2 minus 153 00:09:41,351 --> 00:09:46,610 X&X +3. To write the first term as an equivalent fraction with 154 00:09:46,610 --> 00:09:50,680 this denominator, we multiply top and bottom by X plus three. 155 00:09:50,680 --> 00:09:56,230 So if we multiply top here by X +3 and bottom there by X +3. 156 00:09:56,800 --> 00:09:58,520 Will achieve this fraction. 157 00:09:59,210 --> 00:10:01,613 And this fraction is equivalent to the original 1. 158 00:10:03,550 --> 00:10:06,370 Similarly with the second term. 159 00:10:06,960 --> 00:10:12,630 To achieve a common denominator of 2 minus XX +3. 160 00:10:13,310 --> 00:10:15,152 I need to multiply top and 161 00:10:15,152 --> 00:10:20,968 bottom here. By two minus X, so B times 2 minus X and this 162 00:10:20,968 --> 00:10:24,808 denominator times 2 minus X and that will give me. 163 00:10:25,390 --> 00:10:27,310 B2 minus X at the top. 164 00:10:28,420 --> 00:10:32,012 Now these two fractions have the same denominator, 165 00:10:32,012 --> 00:10:36,053 we can add them together simply by adding the 166 00:10:36,053 --> 00:10:39,645 numerators together, which will give us a multiplied 167 00:10:39,645 --> 00:10:40,992 by X +3. 168 00:10:42,130 --> 00:10:47,807 Plus B multiplied by two minus X. 169 00:10:48,510 --> 00:10:52,450 All divided by the common 170 00:10:52,450 --> 00:10:59,134 denominator. What we're saying is that this fraction we 171 00:10:59,134 --> 00:11:02,699 started with is exactly the 172 00:11:02,699 --> 00:11:05,198 same. As this quantity here. 173 00:11:06,600 --> 00:11:09,484 Now the denominators are already the same. 174 00:11:11,040 --> 00:11:14,784 So if this is the same as that, and the denominators are already 175 00:11:14,784 --> 00:11:18,528 the same, then so too must be the numerators, so we can equate 176 00:11:18,528 --> 00:11:21,984 the numerators if we equate the numerators we can write down X 177 00:11:21,984 --> 00:11:26,750 equals. AX +3. 178 00:11:28,280 --> 00:11:32,288 Plus B2 Minus X. 179 00:11:33,710 --> 00:11:37,989 This is the equation that's going to allow us to calculate 180 00:11:37,989 --> 00:11:39,156 values for A&B. 181 00:11:40,690 --> 00:11:44,122 Now we can find values for A&B in one of two different ways. 182 00:11:44,122 --> 00:11:45,970 The 1st way that I'm going to 183 00:11:45,970 --> 00:11:50,400 look at. Is to substitute specific values in for X. 184 00:11:51,270 --> 00:11:54,558 Remember that this quantity on the left is supposed to be equal 185 00:11:54,558 --> 00:11:58,394 to this on the right for any value of X at all. So in 186 00:11:58,394 --> 00:12:01,408 particular, we can choose any values that we like. That will 187 00:12:01,408 --> 00:12:02,778 make all this look simpler. 188 00:12:03,430 --> 00:12:07,435 And what I'm going to do is I'm going to choose X to have the 189 00:12:07,435 --> 00:12:09,304 value to. Why would I do that? 190 00:12:09,820 --> 00:12:13,744 I choose X to have the value too, because then this second 191 00:12:13,744 --> 00:12:18,322 term will become zero and have 2 - 2, which is zero. Will lose 192 00:12:18,322 --> 00:12:20,875 this term. And we'll be able to 193 00:12:20,875 --> 00:12:24,740 calculate A. So by careful choice of values for X, we can 194 00:12:24,740 --> 00:12:26,246 make this look a lot simpler. 195 00:12:27,600 --> 00:12:28,910 So with X is 2. 196 00:12:29,470 --> 00:12:31,276 On the left will have two. 197 00:12:32,110 --> 00:12:39,078 On the right will have 2 + 3, which is 5 times a. 198 00:12:39,520 --> 00:12:40,790 And this term will vanish. 199 00:12:41,620 --> 00:12:46,339 This gives me a value for a straightaway dividing both sides 200 00:12:46,339 --> 00:12:50,200 by 5. I can write that a is 2/5. 201 00:12:52,240 --> 00:12:54,220 We need to find B. 202 00:12:55,660 --> 00:13:00,910 Now a sensible value that will enable us to find B is to let X 203 00:13:00,910 --> 00:13:03,010 be minus three whi, is that? 204 00:13:03,530 --> 00:13:06,767 Well, if X is minus three, will have minus 3 + 3, which 205 00:13:06,767 --> 00:13:09,257 is zero. And all of this first term will vanish. 206 00:13:10,970 --> 00:13:17,410 And we'll be able to find be so letting XP minus three will have 207 00:13:17,410 --> 00:13:22,930 minus three on the left zero from this term here, and two 208 00:13:22,930 --> 00:13:27,990 minus minus three, which is 2 + 3, which is 55-B. 209 00:13:28,230 --> 00:13:35,838 Dividing both sides by 5 will give us a value for B's 210 00:13:35,838 --> 00:13:38,374 minus three over 5. 211 00:13:39,280 --> 00:13:43,242 So now we know a value for a. We know a value for B. 212 00:13:44,250 --> 00:13:47,610 And we can then proceed to evaluate the integral by 213 00:13:47,610 --> 00:13:48,954 evaluating each of these 214 00:13:48,954 --> 00:13:56,034 separately. Let me write this down again. We want 215 00:13:56,034 --> 00:14:03,564 the integral of X divided by 2 minus XX +3. 216 00:14:03,600 --> 00:14:06,260 With respect to X. 217 00:14:06,260 --> 00:14:09,959 We expressed this algebraic fraction in its partial fraction 218 00:14:09,959 --> 00:14:14,480 in its partial fractions, and we found that a was 2/5. 219 00:14:15,170 --> 00:14:18,818 And be was 220 00:14:18,818 --> 00:14:25,024 minus 3/5. So instead of integrating this original 221 00:14:25,024 --> 00:14:31,393 fraction, what we're going to do now is integrate separately the 222 00:14:31,393 --> 00:14:33,130 two partial fractions. 223 00:14:33,250 --> 00:14:38,080 And will integrate these separately and will do it like 224 00:14:38,080 --> 00:14:42,090 this. In the first integral, we're going to take out the 225 00:14:42,090 --> 00:14:47,314 factor of 2/5. I will be left with the problem of integrating 226 00:14:47,314 --> 00:14:50,226 one over 2 minus X with respect 227 00:14:50,226 --> 00:14:57,052 to X. For the second, we're going to take out 228 00:14:57,052 --> 00:15:03,264 minus 3/5. And integrate one over X +3 with 229 00:15:03,264 --> 00:15:05,340 respect to X. 230 00:15:07,710 --> 00:15:09,887 So the problem of integrating this algebraic 231 00:15:09,887 --> 00:15:12,375 fraction has been split into the problem of 232 00:15:12,375 --> 00:15:14,552 evaluating these two separate integrals and both 233 00:15:14,552 --> 00:15:17,662 of these are simpler than the one we started with. 234 00:15:18,700 --> 00:15:22,600 Let's deal with the second one first. The second one is a 235 00:15:22,600 --> 00:15:26,175 situation where we've got a function at the bottom and it's 236 00:15:26,175 --> 00:15:30,075 derivative at the top. Because we've got X plus three at the 237 00:15:30,075 --> 00:15:34,300 bottom and the derivative of X +3 is just one which appears at 238 00:15:34,300 --> 00:15:37,875 the top. So this just evaluates to minus 3/5 the natural 239 00:15:37,875 --> 00:15:40,800 logarithm of the modulus of what's at the bottom. 240 00:15:42,570 --> 00:15:45,468 We've got the similar situation here, except if you 241 00:15:45,468 --> 00:15:48,366 differentiate the denominator, you get minus one because of 242 00:15:48,366 --> 00:15:52,552 this minus X, so we'd really like a minus one at the top. 243 00:15:53,620 --> 00:15:57,611 And I can adjust my numerator to make it minus one, provided that 244 00:15:57,611 --> 00:16:00,681 I counteract that with putting a minus sign outside there. 245 00:16:01,670 --> 00:16:06,602 So we can write all this as minus 2/5 the natural logarithm 246 00:16:06,602 --> 00:16:12,767 of the modulus of 2 minus X. And of course we need a constant of 247 00:16:12,767 --> 00:16:14,822 integration at the very end. 248 00:16:15,470 --> 00:16:20,534 So that's the result of integrating X over 2 minus XX +3 249 00:16:20,534 --> 00:16:22,222 and the problems finished. 250 00:16:22,920 --> 00:16:28,424 What I'd like to do is just go back a page and just show you an 251 00:16:28,424 --> 00:16:31,864 alternative way of calculating values for the constants A&B in 252 00:16:31,864 --> 00:16:35,648 the partial fractions, and I want us to return to this 253 00:16:35,648 --> 00:16:39,776 equation here that we use to find A&B. Let me write that 254 00:16:39,776 --> 00:16:40,808 equation down again. 255 00:16:41,490 --> 00:16:47,580 X is equal to a X +3. 256 00:16:47,580 --> 00:16:50,700 Plus B2 Minus 257 00:16:50,700 --> 00:16:58,170 X. What I'm going to do is I'm going to 258 00:16:58,170 --> 00:17:01,420 start by removing the brackets. 259 00:17:01,650 --> 00:17:05,310 Will have a multiplied by X 260 00:17:05,310 --> 00:17:10,146 AX. A Times 3 which is 3A. 261 00:17:11,940 --> 00:17:15,258 B times two or two B. 262 00:17:16,130 --> 00:17:19,346 And be times minus X or minus BX. 263 00:17:21,050 --> 00:17:24,768 And then what I'm going to do is, I'm going to collect similar 264 00:17:24,768 --> 00:17:28,200 terms together so you see Ivan Axe here and minus BX there. 265 00:17:28,770 --> 00:17:33,820 So altogether I have a minus B, lots of X. 266 00:17:37,280 --> 00:17:41,739 And we've got 3A Plus 2B here. 267 00:17:44,620 --> 00:17:47,890 We now use this equation to equate coefficients on both 268 00:17:47,890 --> 00:17:52,468 sides. What do we mean by that? Well, what we do is we ask 269 00:17:52,468 --> 00:17:57,046 ourselves how many X terms do we have on the left and match that 270 00:17:57,046 --> 00:18:01,624 with the number of X terms that we have on the right. So you 271 00:18:01,624 --> 00:18:06,529 see, on the left hand side here, if we look at just the ex terms, 272 00:18:06,529 --> 00:18:13,228 there's 1X. On the right we've got a minus B, lots of X. 273 00:18:13,260 --> 00:18:18,060 So we've equated the coefficients of X on both sides. 274 00:18:19,360 --> 00:18:22,948 We can also look at constant terms on both sides. You see the 275 00:18:22,948 --> 00:18:24,880 three A plus 2B is a constant. 276 00:18:25,680 --> 00:18:30,356 There are no constant terms on the left, so if we just look at 277 00:18:30,356 --> 00:18:32,360 constants, there are none on the 278 00:18:32,360 --> 00:18:37,360 left. And on the right there's 3A Plus 2B. 279 00:18:37,360 --> 00:18:40,690 And you'll see what we have. Here are two simultaneous 280 00:18:40,690 --> 00:18:44,353 equations for A&B and if we solve these equations we can 281 00:18:44,353 --> 00:18:48,682 find values for A&B. Let me call that equation one and that one 282 00:18:48,682 --> 00:18:52,678 equation two. What I'm going to do is I'm going to multiply 283 00:18:52,678 --> 00:18:57,673 equation one by two so that will end up with two be so that we 284 00:18:57,673 --> 00:19:01,336 will be able to add these together to eliminate the bees. 285 00:19:01,336 --> 00:19:06,331 So if I take equation one and I multiply it by two, I'll get 2 286 00:19:06,331 --> 00:19:13,538 ones or two. 2A minus 2B. Let's call that equation 3. 287 00:19:14,350 --> 00:19:18,028 If we add equations two and 288 00:19:18,028 --> 00:19:24,622 three together. We've got 0 + 2, which is 2, three, 8 + 2 A which 289 00:19:24,622 --> 00:19:30,750 is 5A and two be added to minus 2B cancels out, so two is 5. In 290 00:19:30,750 --> 00:19:35,729 other words, A is 2 over 5, which is the value we had 291 00:19:35,729 --> 00:19:37,261 earlier on for A. 292 00:19:37,820 --> 00:19:42,580 We can then take this value for A and substitute it in either of 293 00:19:42,580 --> 00:19:46,320 these equations and obtain a value for be. So, for example, 294 00:19:46,320 --> 00:19:48,360 if we substitute in the first 295 00:19:48,360 --> 00:19:54,790 equation. Will find that one equals a, is 2/5 minus B. 296 00:19:56,100 --> 00:20:02,049 Rearranging this B is equal to 2/5 - 1. 297 00:20:02,790 --> 00:20:08,782 And 2/5 - 1 is minus 3/5 the same value as we got before. 298 00:20:09,760 --> 00:20:13,500 So we've seen two ways of finding the values of the 299 00:20:13,500 --> 00:20:17,580 constants A&B. We can substitute specific values for X or we can 300 00:20:17,580 --> 00:20:18,940 equate coefficients on both 301 00:20:18,940 --> 00:20:23,212 sides. Often will need to use a mix of the two methods in order 302 00:20:23,212 --> 00:20:24,920 to find all the constants in a 303 00:20:24,920 --> 00:20:31,405 given problem. Let's have a look at 304 00:20:31,405 --> 00:20:34,456 a definite integral. 305 00:20:35,170 --> 00:20:39,031 Suppose we want to find the integral from X is one to access 306 00:20:39,031 --> 00:20:46,560 2. Of three divided by XX plus one with respect to 307 00:20:46,560 --> 00:20:52,800 X. As before, we examine this integrand and ask ourselves, is 308 00:20:52,800 --> 00:20:55,175 this a proper or improper 309 00:20:55,175 --> 00:20:59,045 fraction? Well, the degree of the denominator is too, because 310 00:20:59,045 --> 00:21:03,010 when we multiply this out, the highest power of X will be 2. 311 00:21:03,620 --> 00:21:09,236 The degree of the numerator is zero, with really 3X to the zero 312 00:21:09,236 --> 00:21:12,692 here, so this is an example of a 313 00:21:12,692 --> 00:21:18,092 proper fraction. On both of these factors are linear 314 00:21:18,092 --> 00:21:23,960 factors. So as before, I'm going to express the integrand. 315 00:21:24,580 --> 00:21:27,044 As the sum of its partial fractions. So let's do that 316 00:21:27,044 --> 00:21:31,040 first of all. 3 divided by XX 317 00:21:31,040 --> 00:21:35,958 plus one. The appropriate form of partial fractions. 318 00:21:36,930 --> 00:21:42,750 Are constant. Over the first linear factor. 319 00:21:42,750 --> 00:21:47,518 Plus another constant over the second linear factor. 320 00:21:49,160 --> 00:21:52,652 And our job now is to try to find values for A&B. 321 00:21:53,720 --> 00:22:00,122 We do this by adding these together as we did before, 322 00:22:00,122 --> 00:22:04,778 common denominator XX plus one in both cases. 323 00:22:06,810 --> 00:22:10,870 To write a over X as an equivalent fraction with 324 00:22:10,870 --> 00:22:14,118 this denominator will need to multiply top and 325 00:22:14,118 --> 00:22:16,148 bottom by X plus one. 326 00:22:18,090 --> 00:22:23,927 To write B over X plus one with this denominator will need to 327 00:22:23,927 --> 00:22:26,621 multiply top and bottom by X. 328 00:22:27,420 --> 00:22:32,840 So now we've given these two fractions a common denominator, 329 00:22:32,840 --> 00:22:37,718 and we add the fractions together by adding the 330 00:22:37,718 --> 00:22:43,301 numerators. I'm putting the result over the common 331 00:22:43,301 --> 00:22:51,224 denominator. So 3 divided by XX plus One is equal to all this. 332 00:22:53,150 --> 00:22:54,805 The denominators are already the 333 00:22:54,805 --> 00:23:00,050 same. So we can equate the numerators that gives us the 334 00:23:00,050 --> 00:23:05,432 equation 3 equals a X plus one plus BX. And this is the 335 00:23:05,432 --> 00:23:09,986 equation we can use to try to find values for A&B. 336 00:23:11,260 --> 00:23:13,636 We could equate coefficients, or we can substitute specific 337 00:23:13,636 --> 00:23:17,332 values for X and what I'm going to do is I'm going to substitute 338 00:23:17,332 --> 00:23:21,028 the value X is not and the reason why I'm picking X is not 339 00:23:21,028 --> 00:23:23,404 is because I recognize straight away that's going to. 340 00:23:23,960 --> 00:23:26,552 Kill off this last term here that'll have gone and will be 341 00:23:26,552 --> 00:23:28,280 able to just find a value for A. 342 00:23:29,280 --> 00:23:34,922 So we substitute X is not on the left, will still have 3. 343 00:23:35,790 --> 00:23:41,601 And on the right we've got not plus one which is one 1A. 344 00:23:41,630 --> 00:23:43,611 Be times not is not so that 345 00:23:43,611 --> 00:23:48,248 goes. So In other words, we've got a value for A and a is 3. 346 00:23:49,640 --> 00:23:54,320 Another sensible value to substitute is X equals minus 347 00:23:54,320 --> 00:23:56,565 one. Why is that a sensible 348 00:23:56,565 --> 00:24:00,294 value? Well, that's a sensible value, because if we put X is 349 00:24:00,294 --> 00:24:03,710 minus one in minus one plus one is zero and will lose this first 350 00:24:03,710 --> 00:24:05,662 term with the A in and will now 351 00:24:05,662 --> 00:24:09,814 be. So putting X is minus one will have 3. 352 00:24:10,360 --> 00:24:14,130 This will become zero and will have be times minus one which is 353 00:24:14,130 --> 00:24:19,700 minus B. So this tells us that B is actually minus three. 354 00:24:20,390 --> 00:24:26,366 So now we know the value of a is going to be 3 and B is going to 355 00:24:26,366 --> 00:24:30,878 be minus three. And the problem of performing this integration 356 00:24:30,878 --> 00:24:34,244 can be solved by integrating these two terms separately. 357 00:24:35,630 --> 00:24:41,306 Let's do that now. I'll write these these terms down again. 358 00:24:41,306 --> 00:24:46,982 We're integrating three over XX plus one with respect to X. 359 00:24:48,180 --> 00:24:52,310 And we've expressed already this as its partial fractions, and 360 00:24:52,310 --> 00:24:56,853 found that we're integrating three over X minus three over X, 361 00:24:56,853 --> 00:25:02,222 plus one with respect to X, and this was a definite integral. It 362 00:25:02,222 --> 00:25:06,352 had limits on, and the limits will one and two. 363 00:25:06,890 --> 00:25:10,256 So now we use partial fractions to change this 364 00:25:10,256 --> 00:25:12,874 algebraic fraction into these two simple integrals. 365 00:25:13,970 --> 00:25:16,088 Now, these are straightforward to finish 366 00:25:16,088 --> 00:25:19,618 because the integral of three over X is just three 367 00:25:19,618 --> 00:25:22,089 natural logarithm of the modulus of X. 368 00:25:25,650 --> 00:25:27,408 The integral of three over X 369 00:25:27,408 --> 00:25:32,745 plus one. Is 3 natural logarithm of the modulus of X plus one? 370 00:25:33,460 --> 00:25:35,800 And there's a minus sign in the middle from that. 371 00:25:36,860 --> 00:25:40,180 This is a definite 372 00:25:40,180 --> 00:25:44,110 integral. So we have square brackets and we write the limits 373 00:25:44,110 --> 00:25:45,660 on the right hand side. 374 00:25:46,350 --> 00:25:48,890 The problem is nearly finished. All we have to do 375 00:25:48,890 --> 00:25:50,160 is substitute the limits in. 376 00:25:51,310 --> 00:25:56,590 Upper limit first when X is 2, will have three natural 377 00:25:56,590 --> 00:25:58,030 log of two. 378 00:25:59,150 --> 00:26:02,918 When X is 2 in here will have 379 00:26:02,918 --> 00:26:06,476 minus three. Natural logarithm of 2 + 1, which is 380 00:26:06,476 --> 00:26:10,727 3. So that's what we get when we put the upper limit in. 381 00:26:11,750 --> 00:26:16,524 When we put the lower limit in, when X is one will have three 382 00:26:16,524 --> 00:26:17,888 natural logarithm of 1. 383 00:26:18,640 --> 00:26:23,398 Minus three natural logarithm of 1 + 1, which is 2. So that's 384 00:26:23,398 --> 00:26:28,156 what we get when we put the lower limiting. And of course we 385 00:26:28,156 --> 00:26:31,450 want to find the difference of these two quantities. 386 00:26:32,300 --> 00:26:36,120 Here you'll notice with three 387 00:26:36,120 --> 00:26:40,634 log 2. And over here there's another three 388 00:26:40,634 --> 00:26:44,274 log, two with A minus and minus, so we're adding 389 00:26:44,274 --> 00:26:47,550 another three log 2. So altogether there will be 390 00:26:47,550 --> 00:26:48,642 6 log 2. 391 00:26:49,840 --> 00:26:53,220 That's minus three log 3. 392 00:26:53,220 --> 00:26:56,668 And the logarithm of 393 00:26:56,668 --> 00:27:00,540 1. Is 0 so that banishes. 394 00:27:01,430 --> 00:27:04,455 Now we could leave the answer like that, although more often 395 00:27:04,455 --> 00:27:07,480 than not would probably use the laws of logarithms to try 396 00:27:07,480 --> 00:27:10,780 to tighten this up a little bit and write it in a 397 00:27:10,780 --> 00:27:13,255 different way. You should be aware that multiplier outside, 398 00:27:13,255 --> 00:27:16,830 like this six, can be put inside as a power, so we can 399 00:27:16,830 --> 00:27:19,580 write this as logarithm of 2 to the power 6. 400 00:27:21,040 --> 00:27:23,960 Subtract again a multiplier outside can move inside as a 401 00:27:23,960 --> 00:27:28,048 power so we can write this as logarithm of 3 to the power 3. 402 00:27:29,310 --> 00:27:32,970 And you'll also be aware from your loss of logarithms that if 403 00:27:32,970 --> 00:27:36,325 we're finding the difference of two logarithms, and we can write 404 00:27:36,325 --> 00:27:41,205 that as the logarithm of 2 to the power 6 / 3 to the power 3. 405 00:27:41,450 --> 00:27:44,030 And that's my final answer. 406 00:27:45,280 --> 00:27:52,309 Let's look at another example in which the denominator contains a 407 00:27:52,309 --> 00:27:57,421 repeated linear factor. Suppose we're interested in evaluating 408 00:27:57,421 --> 00:28:05,089 this integral 1 divided by X minus one all squared X Plus 409 00:28:05,089 --> 00:28:12,118 One, and we want to integrate that with respect to X. 410 00:28:12,990 --> 00:28:14,862 So again, we have a proper 411 00:28:14,862 --> 00:28:19,437 fraction. And there is a linear factor here. X plus one, another 412 00:28:19,437 --> 00:28:23,481 linear factor X minus one. But this is a repeated linear factor 413 00:28:23,481 --> 00:28:24,829 because it occurs twice. 414 00:28:25,530 --> 00:28:30,795 The appropriate form of partial fractions will be these. 415 00:28:31,470 --> 00:28:37,608 We want to constant over the 416 00:28:37,608 --> 00:28:40,677 linear factor X 417 00:28:40,677 --> 00:28:47,190 minus one. We want another constant over the linear 418 00:28:47,190 --> 00:28:52,690 factor repeated X minus one squared. And finally we need 419 00:28:52,690 --> 00:28:58,190 another constant. See over this linear factor X plus one. 420 00:28:58,190 --> 00:29:02,040 And our task is before is to try to find values 421 00:29:02,040 --> 00:29:03,440 for the constants AB&C. 422 00:29:04,470 --> 00:29:09,042 We do that as before, by expressing each of these over a 423 00:29:09,042 --> 00:29:12,471 common denominator and the common denominator that we want 424 00:29:12,471 --> 00:29:15,138 is going to be X minus one. 425 00:29:15,700 --> 00:29:19,740 Squared X plus one. 426 00:29:20,780 --> 00:29:24,560 Now to achieve a common denominator of X minus one 427 00:29:24,560 --> 00:29:29,474 squared X Plus one will need to multiply the top and bottom here 428 00:29:29,474 --> 00:29:35,144 by X minus 1X plus one. So we have a X minus 1X plus one. 429 00:29:35,170 --> 00:29:38,455 To achieve the common denominator in this case will 430 00:29:38,455 --> 00:29:42,105 need to multiply top and bottom by X plus one. 431 00:29:42,620 --> 00:29:46,554 So we'll have a BX plus one. 432 00:29:47,120 --> 00:29:51,152 And finally, in this case, to achieve a denominator of X minus 433 00:29:51,152 --> 00:29:55,856 one squared X Plus one will need to multiply top and bottom by X 434 00:29:55,856 --> 00:29:56,864 minus 1 squared. 435 00:29:57,640 --> 00:30:04,376 Now this fraction here is the same as 436 00:30:04,376 --> 00:30:06,902 this fraction here. 437 00:30:07,300 --> 00:30:10,479 Their denominators are already the same, so we can equate the 438 00:30:10,479 --> 00:30:13,947 numerators. So if we just look at the numerators will have one 439 00:30:13,947 --> 00:30:17,993 on the left is equal to the top line here on the right hand 440 00:30:17,993 --> 00:30:25,160 side. AX minus 1X plus one. 441 00:30:25,290 --> 00:30:29,209 Plus B. X plus one. 442 00:30:30,290 --> 00:30:34,110 Plus C. X minus one all 443 00:30:34,110 --> 00:30:38,720 squared. And now we choose some sensible values for X, so 444 00:30:38,720 --> 00:30:42,427 there's a lot of these terms will drop away. For example, 445 00:30:42,427 --> 00:30:46,471 supposing we pick X equals 1, what's the point of picking X 446 00:30:46,471 --> 00:30:50,178 equals one? Well, if we pick axes one, this first term 447 00:30:50,178 --> 00:30:51,526 vanishes. We lose a. 448 00:30:52,220 --> 00:30:55,832 Also, if we pick X equal to 1, the last term vanish is because 449 00:30:55,832 --> 00:30:59,960 we have a 1 - 1 which is zero and will be just left with the 450 00:30:59,960 --> 00:31:03,930 term involving be. So by letting XP, one will have one. 451 00:31:04,610 --> 00:31:07,634 On the left is equal to 0. 452 00:31:08,200 --> 00:31:12,070 One and one here is 22B. 453 00:31:12,070 --> 00:31:19,090 And the last term vanishes. In other words, B is equal to 1/2. 454 00:31:20,380 --> 00:31:24,982 What's another sensible value to pick for X? Well, if we let X 455 00:31:24,982 --> 00:31:26,044 equal minus one. 456 00:31:26,780 --> 00:31:30,641 X being minus one will mean that this term vanish is minus one 457 00:31:30,641 --> 00:31:31,829 plus One is 0. 458 00:31:32,520 --> 00:31:36,953 This term will vanish minus one, plus one is zero and will be 459 00:31:36,953 --> 00:31:41,386 able to find see. So I'm going to let X be minus one. 460 00:31:41,620 --> 00:31:44,707 Will still have the one on the 461 00:31:44,707 --> 00:31:47,890 left. When X is minus one, this 462 00:31:47,890 --> 00:31:50,110 goes. This goes. 463 00:31:50,680 --> 00:31:55,375 And on the right hand side this term here will have minus 1 - 1 464 00:31:55,375 --> 00:31:59,757 is minus two we square it will get plus four, so will get plus 465 00:31:59,757 --> 00:32:04,999 4C. In other words, see is going to be 1/4. 466 00:32:06,070 --> 00:32:11,110 So we've got be. We've got C and now we need to find a value for 467 00:32:11,110 --> 00:32:15,205 a. Now we can substitute any other value we like in here, so 468 00:32:15,205 --> 00:32:18,985 I'm actually going to pick X equals 0. It's a nice simple 469 00:32:18,985 --> 00:32:23,080 value. Effects is zero, will have one on the left if X is 470 00:32:23,080 --> 00:32:26,230 zero there and there will be left with minus one. 471 00:32:26,790 --> 00:32:32,404 Minus 1 * 1 is minus 1 - 1 times a is minus a. 472 00:32:33,100 --> 00:32:38,840 Thanks is 0 here. Will just left with B Times one which is be. 473 00:32:38,850 --> 00:32:44,422 And if X is 0 here will have minus one squared, which is plus 474 00:32:44,422 --> 00:32:50,392 one plus One Times C Plus C. Now we already know values for B and 475 00:32:50,392 --> 00:32:55,964 for C, so we substitute these in will have one is equal to minus 476 00:32:55,964 --> 00:33:00,342 A plus B which is 1/2 plus C which is 1/4. 477 00:33:01,490 --> 00:33:06,150 So rearranging this will have that a is equal to. 478 00:33:06,730 --> 00:33:12,260 Well, with a half and a quarter, that's 3/4 and the one from this 479 00:33:12,260 --> 00:33:17,790 side over the other side is minus one. 3/4 - 1 is going to 480 00:33:17,790 --> 00:33:18,975 be minus 1/4. 481 00:33:18,980 --> 00:33:25,972 So there we have our values for a for B and for C, so a being 482 00:33:25,972 --> 00:33:29,031 minus 1/4 will go back in there. 483 00:33:29,330 --> 00:33:34,062 Be being a half will go back in here and see being a quarter 484 00:33:34,062 --> 00:33:38,118 will go back in there and then we'll have three separate pieces 485 00:33:38,118 --> 00:33:41,498 of integration to do in order to complete the problem. 486 00:33:42,460 --> 00:33:46,468 Let me write these all down again. I'm going to write the 487 00:33:46,468 --> 00:33:49,474 integral out and write these three separate ones down. 488 00:33:49,490 --> 00:33:56,826 So we had that the integral of one over X minus one squared X 489 00:33:56,826 --> 00:34:03,114 Plus one, all integrated with respect to X, is going to equal. 490 00:34:03,860 --> 00:34:09,940 Minus 1/4 the integral of one over X minus one. 491 00:34:10,610 --> 00:34:17,918 They'll be plus 1/2 integral of one over X minus 1 squared. 492 00:34:18,480 --> 00:34:24,384 And they'll be 1/4 493 00:34:24,384 --> 00:34:27,336 over X 494 00:34:27,336 --> 00:34:33,425 plus one. And we want 495 00:34:33,425 --> 00:34:36,500 to DX in 496 00:34:36,500 --> 00:34:42,033 every term. So we've used partial fractions to split 497 00:34:42,033 --> 00:34:45,823 this integrand into three separate terms, and will try and 498 00:34:45,823 --> 00:34:49,613 finish this off. Now. The first integral straightforward when we 499 00:34:49,613 --> 00:34:54,161 integrate one over X minus one will end up with just the 500 00:34:54,161 --> 00:34:57,572 natural logarithm of the denominator, so we'll end up 501 00:34:57,572 --> 00:35:00,983 with minus 1/4 natural logarithm. The modulus of X 502 00:35:00,983 --> 00:35:06,208 minus one. To integrate this term, we're integrating one over 503 00:35:06,208 --> 00:35:10,618 X minus one squared. Let me just do that separately. 504 00:35:10,770 --> 00:35:16,362 To integrate one over X minus one all squared, we make a 505 00:35:16,362 --> 00:35:20,090 substitution and let you equals X minus one. 506 00:35:20,760 --> 00:35:25,282 Do you then will equal du DX, which is just one times DX, so 507 00:35:25,282 --> 00:35:27,866 do you will be the same as DX? 508 00:35:28,450 --> 00:35:32,239 The integral will become the integral of one over. 509 00:35:32,980 --> 00:35:35,676 X minus one squared will be you squared. 510 00:35:36,820 --> 00:35:38,440 And DX is D. 511 00:35:39,210 --> 00:35:42,666 Now, this is straightforward to finish because this is 512 00:35:42,666 --> 00:35:44,970 integrating you to the minus 2. 513 00:35:45,630 --> 00:35:49,040 Increase the power by one becomes you to the minus 1 514 00:35:49,040 --> 00:35:50,590 divided by the new power. 515 00:35:51,090 --> 00:35:52,806 And add a constant of integration. 516 00:35:54,010 --> 00:35:59,197 So when we do this, integration will end up with minus one over 517 00:35:59,197 --> 00:36:01,480 you. Which is minus one over. 518 00:36:02,250 --> 00:36:04,008 X minus one. 519 00:36:04,660 --> 00:36:10,861 And we can put that back into here now. So the integral half 520 00:36:10,861 --> 00:36:16,108 integral of one over X minus one squared will be 1/2. 521 00:36:16,210 --> 00:36:19,876 All that we've got down here, which is minus one over X minus 522 00:36:19,876 --> 00:36:24,960 one. The constant of integration will add another very end. This 523 00:36:24,960 --> 00:36:27,318 integral is going to be 1/4. 524 00:36:27,890 --> 00:36:31,894 The natural logarithm of the modulus of X Plus One and then a 525 00:36:31,894 --> 00:36:33,742 single constant for all of that. 526 00:36:35,110 --> 00:36:39,491 Let me just tidy up this little a little bit. The two logarithm 527 00:36:39,491 --> 00:36:43,198 terms can be combined. We've got a quarter of this logarithm. 528 00:36:43,198 --> 00:36:46,905 Subtract 1/4 of this logarithm, so together we've got a quarter 529 00:36:46,905 --> 00:36:50,612 the logarithm. If we use the laws of logarithms, we've gotta 530 00:36:50,612 --> 00:36:54,656 log subtracted log. So we want the first term X Plus one 531 00:36:54,656 --> 00:36:56,678 divided by the second term X 532 00:36:56,678 --> 00:37:03,280 minus one. And then this term can be written as minus 1/2. 533 00:37:03,280 --> 00:37:06,060 One over X minus one. 534 00:37:06,060 --> 00:37:09,954 Is a constant of integration at the end, and that's the 535 00:37:09,954 --> 00:37:10,662 integration complete. 536 00:37:11,960 --> 00:37:18,440 Now let's look at an example in which we have an improper 537 00:37:18,440 --> 00:37:21,270 fraction. Suppose we have this 538 00:37:21,270 --> 00:37:27,474 integral. The degree of 539 00:37:27,474 --> 00:37:30,730 the numerator 540 00:37:30,730 --> 00:37:36,332 is 3. The degree of the 541 00:37:36,332 --> 00:37:37,556 denominator is 2. 542 00:37:38,610 --> 00:37:41,620 3 being greater than two means that this is an 543 00:37:41,620 --> 00:37:42,222 improper fraction. 544 00:37:43,330 --> 00:37:46,386 Improper fractions requires special treatment and the first 545 00:37:46,386 --> 00:37:49,824 thing we do is division polynomial division. Now, if 546 00:37:49,824 --> 00:37:53,644 you're not happy with Long Division of polynomials, then I 547 00:37:53,644 --> 00:37:57,464 would suggest that you look again at the video called 548 00:37:57,464 --> 00:38:00,868 polynomial division. When examples like this are done very 549 00:38:00,868 --> 00:38:05,124 thoroughly, but what we want to do is see how many times X minus 550 00:38:05,124 --> 00:38:06,948 X squared minus four will divide 551 00:38:06,948 --> 00:38:13,470 into X cubed. The way we do this polynomial division is we 552 00:38:13,470 --> 00:38:19,630 say how many times does X squared divided into X cubed. 553 00:38:20,330 --> 00:38:24,878 That's like asking How many times X squared will go into X 554 00:38:24,878 --> 00:38:29,047 cubed, and clearly when X squared is divided into X cubed, 555 00:38:29,047 --> 00:38:30,942 the answer is just X. 556 00:38:31,920 --> 00:38:35,652 So X squared goes into X cubed X times and we write 557 00:38:35,652 --> 00:38:36,896 the solution down there. 558 00:38:38,130 --> 00:38:42,186 We take what we have just written down and multiply it by 559 00:38:42,186 --> 00:38:45,566 everything here. So X times X squared is X cubed. 560 00:38:46,130 --> 00:38:49,588 X times minus four is minus 4X. 561 00:38:50,370 --> 00:38:51,878 And then we subtract. 562 00:38:52,690 --> 00:38:55,469 X cubed minus X cubed vanish is. 563 00:38:56,040 --> 00:39:00,385 With no access here, and we're subtracting minus 4X, which is 564 00:39:00,385 --> 00:39:01,570 like adding 4X. 565 00:39:03,110 --> 00:39:08,141 This means that when we divide X squared minus four into X cubed, 566 00:39:08,141 --> 00:39:13,559 we get a whole part X and a remainder 4X. In other words, X 567 00:39:13,559 --> 00:39:18,203 cubed divided by X squared minus four can be written as X. 568 00:39:19,360 --> 00:39:25,096 Plus 4X divided by X squared minus 4. 569 00:39:25,270 --> 00:39:30,242 So in order to tackle this integration, we've done the long 570 00:39:30,242 --> 00:39:34,762 division and we're left with two separate integrals to workout. 571 00:39:35,640 --> 00:39:37,980 Now this one is going to be straightforward. Clearly you're 572 00:39:37,980 --> 00:39:41,022 just integrating X, it's going to be X squared over 2. This is 573 00:39:41,022 --> 00:39:43,596 a bit more problematic. Let's have a look at this again. 574 00:39:44,420 --> 00:39:48,083 This is now a proper fraction where the degree of the 575 00:39:48,083 --> 00:39:51,413 numerator is one with the next to the one here. 576 00:39:52,850 --> 00:39:54,512 The degree of the denominator is 577 00:39:54,512 --> 00:39:56,810 2. So it's a proper fraction. 578 00:39:58,130 --> 00:40:00,990 It looks as though we've got a quadratic factor in the 579 00:40:00,990 --> 00:40:05,690 denominator. But in fact, X squared minus four will 580 00:40:05,690 --> 00:40:10,420 factorize. It's in fact, the difference of two squares. So we 581 00:40:10,420 --> 00:40:16,440 can write 4X over X squared minus four as 4X over X minus 2X 582 00:40:16,440 --> 00:40:20,740 plus two. So the quadratic will actually factorize and you'll 583 00:40:20,740 --> 00:40:23,320 see. Now we've got two linear 584 00:40:23,320 --> 00:40:28,483 factors. We can express this in partial fractions. Let's do 585 00:40:28,483 --> 00:40:35,021 that. We've got 4X over X, minus two X +2, and because each of 586 00:40:35,021 --> 00:40:40,158 these are linear factors, the appropriate form is going to be 587 00:40:40,158 --> 00:40:42,493 a over X minus 2. 588 00:40:43,160 --> 00:40:46,980 Plus B over X +2. 589 00:40:49,520 --> 00:40:54,990 We add these together using a common dumped common denominator 590 00:40:54,990 --> 00:41:02,648 X minus two X +2 and will get a X +2 plus BX minus 591 00:41:02,648 --> 00:41:09,210 2. All over the common denominator, X minus two X +2. 592 00:41:12,310 --> 00:41:18,988 Now the left hand side and the right hand side of the same. The 593 00:41:18,988 --> 00:41:23,758 denominators are already the same, so the numerators must be 594 00:41:23,758 --> 00:41:27,574 the same. 4X must equal a X +2 595 00:41:27,574 --> 00:41:29,790 plus B. 6 - 2. 596 00:41:30,850 --> 00:41:35,708 This is the equation that we can use to find the values for A&B. 597 00:41:37,550 --> 00:41:39,458 Let me write that down again. 598 00:41:39,990 --> 00:41:45,758 We've got 4X is a X +2. 599 00:41:46,440 --> 00:41:51,020 Plus BX minus 2. 600 00:41:52,630 --> 00:41:56,844 What's a sensible value to put in for X? Well, if we choose, X 601 00:41:56,844 --> 00:41:58,951 is 2, will lose the second term. 602 00:41:59,620 --> 00:42:02,728 So X is 2 is a good value to put in here. 603 00:42:03,940 --> 00:42:09,100 Faxes 2 on the left hand side will have 4 * 2, which is 8. 604 00:42:10,070 --> 00:42:15,410 If X is 2, will have 2 + 2, which is 44A. 605 00:42:15,410 --> 00:42:17,940 And this term will vanish. 606 00:42:19,000 --> 00:42:22,200 So 8 equals 4A. 607 00:42:22,380 --> 00:42:27,340 A will equal 2 and that's the value for A. 608 00:42:29,200 --> 00:42:32,684 What's another sensible value to put in for X? Well, if X is 609 00:42:32,684 --> 00:42:36,436 minus two, will have minus 2 + 2, which is zero. Will lose this 610 00:42:36,436 --> 00:42:44,200 term. So if X is minus two, will have minus 8 611 00:42:44,200 --> 00:42:46,180 on the left. 612 00:42:46,190 --> 00:42:47,710 This first term will vanish. 613 00:42:48,490 --> 00:42:54,490 An ex being minus two here will have minus 2 - 2 is minus four 614 00:42:54,490 --> 00:42:55,690 so minus 4B. 615 00:42:55,690 --> 00:43:01,114 So be must also be equal to two. 616 00:43:01,120 --> 00:43:06,160 So now we've got values for A and for be both equal to two. 617 00:43:06,160 --> 00:43:08,680 Let's take us back and see what 618 00:43:08,680 --> 00:43:11,110 that means. It means that when 619 00:43:11,110 --> 00:43:14,736 we express. This quantity 4X over X squared minus 620 00:43:14,736 --> 00:43:17,868 four in partial fractions like this, the values of 621 00:43:17,868 --> 00:43:19,956 A&B are both equal to two. 622 00:43:21,140 --> 00:43:26,740 So the integral were working out is the integral of X +2 over X 623 00:43:26,740 --> 00:43:29,540 minus 2 + 2 over X +2. 624 00:43:29,550 --> 00:43:35,694 So there we are. We've now 625 00:43:35,694 --> 00:43:41,838 got three separate integrals to evaluate, 626 00:43:41,838 --> 00:43:47,982 and it's straightforward to finish this 627 00:43:47,982 --> 00:43:54,540 off. The integral of X is X squared divided by two. 628 00:43:55,650 --> 00:44:02,475 The integral of two over X minus 2 equals 2 natural logarithm of 629 00:44:02,475 --> 00:44:05,625 the modulus of X minus 2. 630 00:44:06,430 --> 00:44:13,934 And the integral of two over X +2 is 2 natural logarithm of X 631 00:44:13,934 --> 00:44:16,614 +2 plus a constant of 632 00:44:16,614 --> 00:44:20,947 integration. If we wanted to do, we could use the laws of 633 00:44:20,947 --> 00:44:23,617 logarithms to combine these log rhythmic terms here, but I'll 634 00:44:23,617 --> 00:44:24,952 leave the answer like that.