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These are the user uploaded subtitles that are being translated: 1 00:00:00,230 --> 00:00:04,650 In this video we're going to be talking about how to use partial fractions to evaluate an integral and 2 00:00:04,650 --> 00:00:10,410 partial fractions is just an integration technique similar to use substitution or integration by parts. 3 00:00:10,410 --> 00:00:14,920 It's just a different way to evaluate an integral based on the kind of function that we have. 4 00:00:15,000 --> 00:00:20,460 Usually when you have a fraction like this when we are five divided by quantity to X plus one times 5 00:00:20,460 --> 00:00:21,930 quantity x minus two. 6 00:00:22,080 --> 00:00:27,130 When you have something like that partial fractions might be a good way to evaluate the integral What 7 00:00:27,150 --> 00:00:33,150 part of fractions allows us to do is to simplify a fraction like this one that has factors in its denominator 8 00:00:33,240 --> 00:00:37,910 such that we come up with a function that's easier to integrate than the function that we've been given. 9 00:00:37,920 --> 00:00:42,870 So all we want to do is recognize first of all what kind of factors we have in our denominator in the 10 00:00:42,870 --> 00:00:44,280 denominator of this fraction. 11 00:00:44,280 --> 00:00:47,700 We have two factors One factor to X plus one. 12 00:00:47,700 --> 00:00:49,510 The other factor is X minus 2. 13 00:00:49,590 --> 00:00:53,680 So we need to first distinguish between linear factors and quadratic factors. 14 00:00:53,700 --> 00:00:58,260 These are both linear factors because this is actually 2 x to the first power. 15 00:00:58,260 --> 00:01:03,650 And this is X to the first power in other words we have first degree X variables with X to the first. 16 00:01:03,690 --> 00:01:09,630 So they're linear factors if you have X squared or a larger degree exponent on the x variable you're 17 00:01:09,630 --> 00:01:14,220 looking at a quadratic factor linear factors are going to be easier for us to handle than quadratic 18 00:01:14,220 --> 00:01:14,910 vectors. 19 00:01:14,910 --> 00:01:19,710 The other thing we look at is whether or not we have distinct factors or repeated factors. 20 00:01:19,740 --> 00:01:25,320 In other words do we have multiple of the same factor or all of our factors different in this case 2 21 00:01:25,320 --> 00:01:27,820 x plus one is different than X minus 2. 22 00:01:27,840 --> 00:01:30,150 So these two factors are distinct from one another. 23 00:01:30,150 --> 00:01:31,090 They're not the same. 24 00:01:31,140 --> 00:01:33,690 So we have distinct linear factors. 25 00:01:33,690 --> 00:01:38,310 And when that's the case here's what your partial fractions decomposition is going to look like you're 26 00:01:38,310 --> 00:01:43,710 going to keep the fraction exactly as it is we're going to put that on the left hand side of a new equation 27 00:01:43,710 --> 00:01:44,660 that we're going to create. 28 00:01:44,820 --> 00:01:51,570 So we're going to say five divided by quantity to X plus one times quantity x minus two so exactly the 29 00:01:51,570 --> 00:01:53,140 same fraction that we started with. 30 00:01:53,220 --> 00:01:57,840 Then on the right hand side we're going to put each of the linear factors in the denominator of its 31 00:01:57,900 --> 00:01:58,730 own fraction. 32 00:01:58,740 --> 00:02:05,400 So 2 x plus 1 is going to go in the denominator of one fraction and X minus two is going to go in the 33 00:02:05,400 --> 00:02:10,650 denominator of another fraction and then because these are linear factors we're going to put a single 34 00:02:10,680 --> 00:02:13,080 constant in the numerator of each fraction. 35 00:02:13,080 --> 00:02:14,970 So we'll call this constant a. 36 00:02:15,150 --> 00:02:18,940 And then because we've already used a we'll call this constant B. 37 00:02:18,960 --> 00:02:27,120 Now this right hand side over here this is what we're going to use instead of the original fraction 38 00:02:27,120 --> 00:02:27,810 that we were given. 39 00:02:27,810 --> 00:02:33,270 So we're going to take this value here and we're going to replace this original fraction with this right 40 00:02:33,270 --> 00:02:33,960 hand side. 41 00:02:33,960 --> 00:02:39,780 The only thing we have to do first is solve for values of A and B the way that we're going to find values 42 00:02:39,780 --> 00:02:45,700 for A and B is by multiplying both sides of this equation by the denominator from the left hand side. 43 00:02:45,720 --> 00:02:53,880 So we're going to multiply everything by quantity to X plus one times quantity x minus two this denominator 44 00:02:53,880 --> 00:02:55,150 here from the left hand side. 45 00:02:55,320 --> 00:03:00,360 When we do that we're gonna get this entire denominator to cancel so we're just going to be left with 46 00:03:00,360 --> 00:03:03,980 five on the left hand side on the right hand side. 47 00:03:04,110 --> 00:03:08,800 We'll get this to X plus 1 in the denominator to cancel with this 2 x plus 1. 48 00:03:08,970 --> 00:03:11,190 Leaving us with just the factor of x minus 2. 49 00:03:11,190 --> 00:03:14,880 So we're going to get a times quantity x minus 2. 50 00:03:14,940 --> 00:03:19,530 And then when we multiply this by the second fraction here with B we're going to get X minus 2 in the 51 00:03:19,530 --> 00:03:26,770 denominator to cancel with this X minus 2 in the numerator and we're going to get B times 2 x plus 1. 52 00:03:26,820 --> 00:03:30,700 Now there's a couple ways that we can solve for the values of A and B at this point. 53 00:03:30,810 --> 00:03:35,550 We can multiply everything out on the right hand side collect like terms and then use the method of 54 00:03:35,610 --> 00:03:37,050 undetermined coefficients. 55 00:03:37,170 --> 00:03:42,270 But when the facts are this simple we can use another method we can say if x is equal to 2 then inside 56 00:03:42,270 --> 00:03:47,010 these parentheses here we're going to get to minus 2 or 0 0 times a is 0. 57 00:03:47,010 --> 00:03:51,640 In other words setting X equal to two is going to get rid of this constant. 58 00:03:51,660 --> 00:03:56,060 A It's going to get rid of the variable A and we're just going to be left with B. 59 00:03:56,070 --> 00:03:57,960 So we'll be able to solve for a value of be. 60 00:03:57,960 --> 00:04:03,140 So you want to pick the value of x that's going to make the value inside the parentheses 0. 61 00:04:03,150 --> 00:04:11,660 So here if we say x is equal to 2 so 4 x equals to what we can get is five is equal to. 62 00:04:11,670 --> 00:04:14,850 Here we get to minus 2 or zero zero times a is zero. 63 00:04:14,850 --> 00:04:16,230 So that term disappears. 64 00:04:16,230 --> 00:04:22,020 Over here we're going to get B times 2 times 2 is 4 4 plus 1 is 5. 65 00:04:22,020 --> 00:04:29,010 So we're going to get the times 5 If we divide both sides by 5 We're going to get B is equal to 1. 66 00:04:29,010 --> 00:04:30,270 Now we can do the same thing. 67 00:04:30,300 --> 00:04:31,690 But to solve for a. 68 00:04:31,710 --> 00:04:34,750 We just need to make this 2 x plus 1 equal to zero. 69 00:04:34,770 --> 00:04:39,310 Because if we make this equal to zero then we multiply it by B we'll get B to disappear. 70 00:04:39,420 --> 00:04:41,580 So how do we make 2 x plus one equals zero. 71 00:04:41,580 --> 00:04:47,220 Well this one's a little more complicated but we can just say to X plus one equals zero and solve for 72 00:04:47,220 --> 00:04:52,980 x so 2 x is equal to negative 1 x is equal to negative 1 1/2. 73 00:04:52,980 --> 00:05:03,300 So if we say for X equals negative 1 half what we get then is 5 is go to a multiplied by negative one 74 00:05:03,300 --> 00:05:06,060 half minus two is a negative 5 1/2. 75 00:05:06,060 --> 00:05:12,030 So we get negative five halves here we're going to get plus B when we take negative one having plug 76 00:05:12,030 --> 00:05:17,670 it in here for X. We're going to get negative 1 1/2 times 2 is negative one negative one plus one is 77 00:05:17,670 --> 00:05:19,980 0 0 times b is zero. 78 00:05:19,980 --> 00:05:21,930 So this whole term disappears. 79 00:05:21,930 --> 00:05:28,760 Now if we multiply here both sides by 2 what we get is 10 is equal to negative five. 80 00:05:28,950 --> 00:05:34,520 And then if we divide both sides by negative 5 we get a is equal to negative 2. 81 00:05:34,530 --> 00:05:39,320 So now we have values for a and b we can plug them into our partial fraction decomposition. 82 00:05:39,330 --> 00:05:43,870 So we'll say a is negative 2 and we'll say B is going to be 1. 83 00:05:43,950 --> 00:05:48,180 And now we can take this value and replace the original fraction. 84 00:05:48,240 --> 00:05:53,280 So instead of taking the integral of this original fraction Instead we'll take the integral of negative 85 00:05:53,280 --> 00:06:00,930 2 over 2 x plus one plus one over X minus 2 D x. 86 00:06:01,050 --> 00:06:03,930 And now this is an integral we can easily evaluate. 87 00:06:03,960 --> 00:06:10,610 Remember we know that the integral of 1 divided by X is the natural log of the value of x plus C. 88 00:06:10,740 --> 00:06:16,020 In other words and we want to take the integral of 1 divided by something whatever this is here the 89 00:06:16,080 --> 00:06:21,370 integral of that is going to be natural log of the absolute value of that denominator plus C. 90 00:06:21,480 --> 00:06:26,160 The only thing we have to remember is that we have to apply chain rule so we'll have to divide this 91 00:06:26,400 --> 00:06:28,730 by the derivative of x. 92 00:06:28,770 --> 00:06:30,180 So here's what that looks like. 93 00:06:30,360 --> 00:06:35,070 We can separate these two fractions into their own integrals and we can pull out this constant here. 94 00:06:35,070 --> 00:06:43,410 So I end up with is negative two times the integral of one divided by two x plus 1 D x plus the integral 95 00:06:43,410 --> 00:06:47,230 of 1 over X minus 2 D x. 96 00:06:47,340 --> 00:06:52,650 Now when we take the integral we can take this negative to put that out in front so negative 2 then 97 00:06:52,650 --> 00:06:56,940 we'll say the natural log of the absolute value of the denominators of the natural log of the absolute 98 00:06:56,940 --> 00:07:00,270 value of the denominator to X plus 1. 99 00:07:00,270 --> 00:07:03,750 But then we have to divide it by the derivative of the denominator. 100 00:07:03,960 --> 00:07:09,550 So our denominator is 2 x plus 1 the derivative of 2 x plus 1 is just 2. 101 00:07:09,600 --> 00:07:12,700 So we have to divide by 2. 102 00:07:12,960 --> 00:07:18,750 Then for this integral here the second integral will say plus the natural log of the absolute value 103 00:07:18,810 --> 00:07:20,470 of x minus 2. 104 00:07:20,670 --> 00:07:23,370 But now we have to divide by the derivative of the denominator. 105 00:07:23,370 --> 00:07:25,430 So the denominator is X minus 2. 106 00:07:25,650 --> 00:07:28,370 The derivative of X minus 2 is just 1. 107 00:07:28,380 --> 00:07:33,120 So we have to divide this by 1 but of course dividing by 1 doesn't change anything so we don't actually 108 00:07:33,120 --> 00:07:34,160 have to write that in. 109 00:07:34,260 --> 00:07:37,680 And then we just put plus C to account for that constant of integration. 110 00:07:37,680 --> 00:07:41,340 So then here we can cancel the two in the numerator with the two in the denominator. 111 00:07:41,340 --> 00:07:45,690 Leave me with just negative natural log of the value of 2 x plus 1. 112 00:07:45,750 --> 00:07:52,560 If we reorder these terms we get natural log absolute value X minus 2 minus natural Lague absolute value 113 00:07:52,560 --> 00:07:55,860 of 2 x plus 1 plus C. 114 00:07:56,010 --> 00:08:02,100 And remember when you have natural log of a minus natural log of B. 115 00:08:02,130 --> 00:08:06,850 That's going to be equal to the natural log of a divided by B. 116 00:08:06,870 --> 00:08:12,360 So what we can say here is we can say that this is equal to the natural log of the absolute value of 117 00:08:12,450 --> 00:08:13,380 this first value. 118 00:08:13,380 --> 00:08:14,470 Here this first argument. 119 00:08:14,490 --> 00:08:22,470 X minus 2 divided by the second argument to X plus 1 and then we can add our constant integration. 120 00:08:22,470 --> 00:08:24,540 So this then is our final answer. 121 00:08:24,540 --> 00:08:30,060 This is the integral of the original function which we found using a partial fraction decomposition 122 00:08:30,330 --> 00:08:32,350 with distinct linear factors. 13602