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These are the user uploaded subtitles that are being translated: 1 00:00:00,840 --> 00:00:05,280 You're going to be talking about how you use partial fractions to evaluate an indefinite integral. 2 00:00:05,520 --> 00:00:10,620 And in this particular problem we've been given the integral of DX divided by the quantity x squared 3 00:00:10,650 --> 00:00:15,860 plus X and we've been asked to use partial fractions to evaluate this integral. 4 00:00:15,870 --> 00:00:20,700 Now as you get more and more practice with evaluating integrals it'll become more and more clear to 5 00:00:20,700 --> 00:00:25,950 you which integrals are good for partial fractions which are good for integration by parts and which 6 00:00:25,950 --> 00:00:32,130 are good for you substitution or maybe even a different method if you need any method other than just 7 00:00:32,220 --> 00:00:38,520 evaluating the integral directly one give away for partial fractions is you're looking for a rational 8 00:00:38,520 --> 00:00:39,330 function. 9 00:00:39,420 --> 00:00:45,270 Remember that by a rational function we mean a fraction where both the numerator and denominator are 10 00:00:45,300 --> 00:00:46,450 polynomials. 11 00:00:46,500 --> 00:00:50,370 And in this case we could take the DX out of our Fraction Right. 12 00:00:50,370 --> 00:00:56,010 DX is just essentially multiply by 1 over x squared plus x. 13 00:00:56,060 --> 00:01:02,040 So we have D X here and now it's really clear that we have a rational function because one is this constant 14 00:01:02,040 --> 00:01:04,040 value and x squared plus x. 15 00:01:04,050 --> 00:01:09,120 Both of these functions here are polynomials and we have one polynomial in the numerator one in the 16 00:01:09,120 --> 00:01:09,980 denominator. 17 00:01:10,020 --> 00:01:11,190 That's a rational function. 18 00:01:11,190 --> 00:01:15,590 So you might go right to partial fractions as a way to evaluate this integral. 19 00:01:15,630 --> 00:01:21,360 When you first look at it remember that partial fractions is just another way to simplify an integral 20 00:01:21,360 --> 00:01:26,880 that you can't evaluate directly in the same way that you can use integration by parts and you substitution 21 00:01:27,240 --> 00:01:32,410 to simplify an integral you can't evaluate directly the way that partial fractions works. 22 00:01:32,430 --> 00:01:40,200 To simplify these integrals we're going to basically decompose this fraction into smaller more manageable 23 00:01:40,200 --> 00:01:40,740 fractions. 24 00:01:40,740 --> 00:01:42,180 That's what it allows us to do. 25 00:01:42,330 --> 00:01:44,430 And it makes sense with the name partial fractions. 26 00:01:44,430 --> 00:01:50,850 Basically we're going to split this fraction into different partial fractions that together would make 27 00:01:50,850 --> 00:01:57,480 up this singular fraction but which separately are easier for us to evaluate So we're just trying to 28 00:01:57,480 --> 00:02:00,100 figure out how we can pull it apart. 29 00:02:00,240 --> 00:02:03,740 The person we want to do is make sure that we have a proper function. 30 00:02:03,810 --> 00:02:07,430 You always need to do that before you start your partial fractions decomposition. 31 00:02:07,500 --> 00:02:15,330 And what we mean by a proper function is the degree of the denominator must be greater than the degree 32 00:02:15,330 --> 00:02:16,460 of the numerator. 33 00:02:16,470 --> 00:02:19,630 Remember that the degree of a function like this. 34 00:02:19,650 --> 00:02:25,550 We talk about as the largest exponent on whatever you know x variable we have here. 35 00:02:25,710 --> 00:02:33,210 So in our denominator year we have x squared and essentially X to the first power here the largest exponent 36 00:02:33,240 --> 00:02:35,100 there is this too. 37 00:02:35,220 --> 00:02:41,310 So the degree of our denominator is to it's a second degree polynomial here in the denominator in the 38 00:02:41,310 --> 00:02:41,790 numerator. 39 00:02:41,790 --> 00:02:48,390 We don't even have an x variable so it's essentially X to the zero power or a 0 degree polynomial in 40 00:02:48,390 --> 00:02:49,140 the numerator. 41 00:02:49,140 --> 00:02:54,370 So in this case the degree of the denominator is larger than the degree of the numerator and therefore 42 00:02:54,370 --> 00:02:54,450 it. 43 00:02:54,450 --> 00:02:59,560 This is a proper rational function and we can proceed with our partial fractions see composition. 44 00:02:59,640 --> 00:03:03,430 If you don't have that situation you may need to do one of two things. 45 00:03:03,450 --> 00:03:10,260 Either make some kind of rationalizing substitution or perform a polynomial long division and I have 46 00:03:10,260 --> 00:03:16,260 videos about both of those methods so if you have that situation where your function is not proper then 47 00:03:16,290 --> 00:03:20,640 check out one of those videos to learn more about how to make it proper. 48 00:03:20,670 --> 00:03:26,700 But once we have it proper what we want to do is factor the denominator as completely as possible. 49 00:03:26,730 --> 00:03:31,200 We're going to take this function outside of the integral for now and we're going to factor the denominator 50 00:03:31,200 --> 00:03:32,530 as much as we can. 51 00:03:32,580 --> 00:03:38,520 So the denominator here we have x squared plus X we can factor out an X from that and what we'll get 52 00:03:38,550 --> 00:03:42,000 is X times x plus 1. 53 00:03:42,180 --> 00:03:45,100 And you want to factor as completely as you can. 54 00:03:45,210 --> 00:03:49,490 The more complicated here your polynomial may be the more factors you can have. 55 00:03:49,500 --> 00:03:55,650 But you want to get to the point where your factors are irreducible you can't factor them any further 56 00:03:55,720 --> 00:04:00,660 than you already have and that's what we've got here now at this point you want to look at the factors 57 00:04:00,660 --> 00:04:05,460 that you have and your denominator and there's a couple situations that can arise here. 58 00:04:05,580 --> 00:04:12,860 You can have either linear and or quadratic factors a linear factor is a factor whose degree is one. 59 00:04:12,870 --> 00:04:14,970 Right here we have two factors. 60 00:04:14,970 --> 00:04:23,010 We have X and we have X plus 1 in our factor here X we have X to the first power and an R factor X plus 61 00:04:23,010 --> 00:04:23,370 1. 62 00:04:23,370 --> 00:04:25,360 We have X to the first plus 1. 63 00:04:25,590 --> 00:04:30,660 These are both first degree factors and therefore their linear factors. 64 00:04:30,660 --> 00:04:36,750 If we had X squared for this first factor here that would be a quadratic factor anything x squared x 65 00:04:36,750 --> 00:04:40,950 cubed X to the fourth anything greater than 1 is a quadratic factor. 66 00:04:41,070 --> 00:04:46,800 And I have videos about how to do partial fractions decomposition with quadratic factors so you can 67 00:04:46,800 --> 00:04:48,940 check that out in a different video. 68 00:04:49,050 --> 00:04:51,640 Here we're only going to talk about linear factors. 69 00:04:51,720 --> 00:04:56,600 You can also have one of two situations with either of those linear or quadratic factors. 70 00:04:56,640 --> 00:05:03,670 You can have distinct factors and or repeated fact what we mean by that is if we were to have for example 71 00:05:03,820 --> 00:05:13,420 if our denominator was X times x plus 1 squared then this X plus 1 factor here would be a repeated factor 72 00:05:13,420 --> 00:05:19,390 because essentially our denominator would be x times the quantity x plus 1 times the quantity x plus 73 00:05:19,390 --> 00:05:22,550 1 because we had quantity x plus 1 squared. 74 00:05:22,840 --> 00:05:25,960 And you can see X plus 1 here is repeated. 75 00:05:25,960 --> 00:05:30,240 So we would call that a repeated factor whereas this x factor here. 76 00:05:30,460 --> 00:05:32,790 This factor only occurs one time. 77 00:05:32,800 --> 00:05:35,540 So we call that a distinct factor. 78 00:05:35,560 --> 00:05:40,360 I have other videos about how to deal with repeated factors so if you ever needed factors go check out 79 00:05:40,360 --> 00:05:41,080 one of those. 80 00:05:41,200 --> 00:05:46,650 But essentially what you can see here is that we're dealing with distinct linear factors. 81 00:05:46,690 --> 00:05:53,050 You can have distinct linear factors distinct quadratic factors repeated linear factors and repeated 82 00:05:53,050 --> 00:05:56,710 quadratic factors so all of those situations are going to be slightly different. 83 00:05:56,710 --> 00:05:59,530 Here we're dealing with distinct linear factors. 84 00:05:59,530 --> 00:06:04,990 So in this case when we have distinct linear factors we're going to set this equal to our partial fractions 85 00:06:04,990 --> 00:06:06,880 the composition on the right hand side. 86 00:06:07,120 --> 00:06:13,270 And when you have a linear factor you just want to put a single variable A as the numerator of that 87 00:06:13,270 --> 00:06:13,750 fraction. 88 00:06:13,750 --> 00:06:20,230 So we're going to have a over our first linear factor X and for our second factor we're going to use 89 00:06:20,230 --> 00:06:22,080 the next variable which would be B. 90 00:06:22,090 --> 00:06:26,830 So we're going to have B over our next linear factor X plus 1. 91 00:06:26,960 --> 00:06:31,210 Now a and b are going to change if you have quadratic factors if you have repeated factors you're going 92 00:06:31,210 --> 00:06:32,700 to need to change this a little bit. 93 00:06:32,830 --> 00:06:37,810 But when it's just linear factors and they're all distinct from one another they're not repeat in you 94 00:06:37,810 --> 00:06:44,050 just take each factor you put it as the denominator of its own fraction and you put a variable in the 95 00:06:44,050 --> 00:06:48,700 numerator of each fraction and you never repeat these variables so if we had a third factor we'd have 96 00:06:48,700 --> 00:06:52,870 another fraction you know plus C over X plus 2 or something like that. 97 00:06:52,870 --> 00:06:57,220 So this is distinct linear factors and once you have it set up like this here's what we're going to 98 00:06:57,220 --> 00:07:00,480 do and this is where things start to get a little interesting. 99 00:07:00,700 --> 00:07:05,800 We're going to multiply both sides of this function by the denominator from the left hand side x times 100 00:07:05,800 --> 00:07:10,670 the quantity x plus 1 x times the quantity x plus 1. 101 00:07:10,840 --> 00:07:15,860 When we do that we're going to multiply each term on both the left and right hand side by this value 102 00:07:16,480 --> 00:07:18,140 that's going to get on the left hand side. 103 00:07:18,160 --> 00:07:24,100 This denominator to cancel completely and all we're left with is 1 in the numerator over here on the 104 00:07:24,100 --> 00:07:30,790 left hand side on the right hand side when we multiply X times x plus 1 by this over X. We're going 105 00:07:30,790 --> 00:07:39,760 to get this x denominator here to cancel with this x and we're just left with a times x plus 1 for here 106 00:07:39,790 --> 00:07:44,470 b over X plus 1 the X plus 1 is going to cancel with this X plus 1 term. 107 00:07:44,710 --> 00:07:48,200 And we're just left with B times x. 108 00:07:48,340 --> 00:07:53,260 Once we've done this we want to go ahead and multiply out everything on the right hand side completely. 109 00:07:53,290 --> 00:07:57,460 So we'll have a x plus a plus b x. 110 00:07:57,490 --> 00:08:03,910 When we distribute that a across the X plus 1 then we want to group terms that have the same degree 111 00:08:03,910 --> 00:08:04,390 together. 112 00:08:04,390 --> 00:08:09,480 So we have here X to the first power and x to the first power which means those are going to go together. 113 00:08:09,700 --> 00:08:15,190 All of our constants if we have just a lone A or a loan B are constants are going to go together. 114 00:08:15,490 --> 00:08:20,440 If we had X squared values over here on the right they would all go together so we're going to say one 115 00:08:20,440 --> 00:08:21,940 is equal to. 116 00:08:22,450 --> 00:08:29,590 And we'll put them in parentheses so x plus be X like this and then plus a is our only constant so it 117 00:08:29,590 --> 00:08:31,230 goes over here by itself. 118 00:08:31,240 --> 00:08:36,700 Once you've done that you want to factor the X out from each of these values here so we'll say one is 119 00:08:36,700 --> 00:08:37,520 equal to. 120 00:08:37,600 --> 00:08:44,110 We're going to factor the X out of this and we'll get a plus b times X. We just pulled the X out and 121 00:08:44,110 --> 00:08:45,280 then we have plus a. 122 00:08:45,460 --> 00:08:51,610 Now the reason that we're doing this is because we're going to end up comparing coefficients on the 123 00:08:51,610 --> 00:08:57,370 left and right hand sides to create simultaneous equations that we can use to solve for our constants 124 00:08:57,490 --> 00:08:58,680 A and B. 125 00:08:58,750 --> 00:09:04,270 So what we have to do now is equate coefficient on the left hand side to those on the right hand side. 126 00:09:04,480 --> 00:09:10,990 And we have to think of our left hand side over here as 0 x plus 1. 127 00:09:11,020 --> 00:09:14,180 Right we haven't changed anything because 0 times X is 0. 128 00:09:14,200 --> 00:09:17,160 So we have to add anything actually to the left hand side. 129 00:09:17,470 --> 00:09:23,740 What we've done though is allowed ourselves the ability to identify that the coefficient of x on the 130 00:09:23,740 --> 00:09:26,130 left hand side is zero. 131 00:09:26,260 --> 00:09:31,820 The coefficient of x on the right hand side is a plus b right here. 132 00:09:32,200 --> 00:09:38,950 And then we equate our constants the constant on the left hand side is this positive one constant on 133 00:09:38,950 --> 00:09:42,100 the right hand side is this positive a right here. 134 00:09:42,400 --> 00:09:50,320 So what we do now is we equate those so we say zero from the left hand side is equal to a plus b and 135 00:09:50,320 --> 00:09:55,160 we say that one from the left hand side is equal to a. 136 00:09:55,180 --> 00:10:01,880 Now we have two equations and we can use the two of these to solve for our constants A and B Well conveniently 137 00:10:02,180 --> 00:10:07,890 We already have a value for a we know that a is equal to one from the second equation here. 138 00:10:07,910 --> 00:10:14,690 Now we can go ahead and plug in one for a to this other equation so we'll get zero is equal to a plug 139 00:10:14,690 --> 00:10:15,960 in 1 for a. 140 00:10:16,100 --> 00:10:20,890 And then we'll get plus B when we subtract 1 from both sides. 141 00:10:20,900 --> 00:10:24,730 We can see that we get B is equal to negative 1. 142 00:10:24,950 --> 00:10:30,920 And now we have a value for both a and b we've solved for our constants at this point we plug them back 143 00:10:31,010 --> 00:10:33,710 in to our partial fractions decomposition. 144 00:10:33,860 --> 00:10:40,040 This decomposition rate here that we had from before and we're going to be taking the integral of this 145 00:10:40,050 --> 00:10:45,410 decomposition instead of our original integral that we started with. 146 00:10:45,470 --> 00:10:52,370 So we're going to get the integral of a which we know is positive 1 divided by X from our decomposition 147 00:10:52,370 --> 00:10:58,610 before then we have plus B but we know that B is negative ones who are actually going to get minus 1 148 00:10:59,180 --> 00:11:04,580 divided by this original denominator here X plus 1 from our decomposition. 149 00:11:04,580 --> 00:11:10,310 And this is our new integral and as you can see these two fractions individually are much simpler to 150 00:11:10,310 --> 00:11:12,190 integrate than our original. 151 00:11:12,190 --> 00:11:18,140 Fraction one over x squared plus X and that's y partial fractions is so useful. 152 00:11:18,230 --> 00:11:24,080 So taking the integral of both of these terms individually you know that the integral of one over X 153 00:11:24,170 --> 00:11:29,520 is natural log of the absolute value of x or any natural log the absolute value of x. 154 00:11:29,900 --> 00:11:35,300 Then here for our second fraction we'll get minus because we have this negative sign here minus natural 155 00:11:35,300 --> 00:11:40,050 log of the absolute value of our denominator X plus 1. 156 00:11:40,340 --> 00:11:46,350 But then we need to apply chain rule and divide by the derivative of this X plus one value here. 157 00:11:46,490 --> 00:11:49,550 Well the derivative of X plus 1 is just one. 158 00:11:49,910 --> 00:11:52,630 So we divide by 1 it doesn't really change anything. 159 00:11:52,790 --> 00:11:56,740 Then we just need to remember to add c to account for the constant of integration. 160 00:11:56,930 --> 00:12:01,400 At this point we have our final answer except that we want to simplify as much as we can remember that 161 00:12:01,400 --> 00:12:09,350 we have a property of natural locks that tells us that natural log of a minus natural log of B is equal 162 00:12:09,350 --> 00:12:14,860 to natural log of a divided by B and that's the situation we have here. 163 00:12:15,050 --> 00:12:19,970 So instead of leaving the separate we can combine them and say that the simplification of our final 164 00:12:19,970 --> 00:12:27,650 answer is the natural log of X divided by X plus 1 and we keep our absolute value bars because we have 165 00:12:27,650 --> 00:12:33,900 those on both natural logs plus seeing is the integral of our original function. 166 00:12:33,920 --> 00:12:34,480 So that's it. 167 00:12:34,490 --> 00:12:35,590 That's our final answer. 168 00:12:35,600 --> 00:12:39,170 That's how you use partial fractions to evaluate an indefinite integral. 19321