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These are the user uploaded subtitles that are being translated: 1 00:00:00,240 --> 00:00:04,410 Today we're going to talk about how you use partial fractions to integrate a rational function with 2 00:00:04,410 --> 00:00:10,770 distinct quadratic factors in its denominator to complete this problem will set up the decomposition 3 00:00:10,800 --> 00:00:14,530 and solve for constants and then integrate the decomposition. 4 00:00:14,790 --> 00:00:19,470 In this particular problem we've been asked to use partial fractions to take the integral of the rational 5 00:00:19,470 --> 00:00:21,110 function that I've written here. 6 00:00:21,120 --> 00:00:25,740 Our first step with any partial fractions problem is to ensure that the degree of the denominator is 7 00:00:25,740 --> 00:00:30,900 greater than the degree of the numerator meaning that the largest exponent in the denominator is greater 8 00:00:30,900 --> 00:00:33,300 than the largest exponent in the numerator. 9 00:00:33,300 --> 00:00:36,300 In this case that's true so we can move on to the next step. 10 00:00:36,300 --> 00:00:42,150 If it weren't true for example if we add X to the fourth in the numerator and x cubed in the denominator 11 00:00:42,450 --> 00:00:47,450 we need to use polynomial long division to simplify our rational function. 12 00:00:47,460 --> 00:00:52,740 But in this case we can go ahead and move onto the next step and our next step will be to factor the 13 00:00:52,800 --> 00:00:57,380 denominator of our rational function as completely as possible. 14 00:00:57,630 --> 00:01:07,380 So the denominator will factor into the quantity x squared plus four times the quantity x squared plus 15 00:01:07,380 --> 00:01:08,100 1. 16 00:01:08,130 --> 00:01:11,480 We always want to factor as completely as possible in this case. 17 00:01:11,490 --> 00:01:13,630 This is as far as we can go. 18 00:01:13,710 --> 00:01:19,560 Once we've done map we want to set this equal to our partial fractions the composition which will put 19 00:01:19,560 --> 00:01:21,460 over here on the right hand side. 20 00:01:21,570 --> 00:01:23,290 So a partial fraction decomposition. 21 00:01:23,300 --> 00:01:28,100 What we do is we separate each term in the denominator. 22 00:01:28,100 --> 00:01:31,610 Each factor in the denominator into its own fraction. 23 00:01:31,620 --> 00:01:33,260 Over here on the right hand side. 24 00:01:33,330 --> 00:01:41,400 So we'll put the first factor X squared plus four into its own fraction and we'll put the second factor 25 00:01:41,550 --> 00:01:45,810 X squared plus 1 into its own fraction like this. 26 00:01:45,810 --> 00:01:50,400 Now the question is what do we need to put in the numerator of each of these fractions Well when we're 27 00:01:50,400 --> 00:01:56,010 dealing with quadratic factors meaning that the degree of the factor is 2 or greater. 28 00:01:56,010 --> 00:02:01,530 So we have x squared plus for the degree is 2 or greater. 29 00:02:01,530 --> 00:02:03,490 That means that it's a quadratic factor. 30 00:02:03,600 --> 00:02:06,110 If it's X to the first power or just X.. 31 00:02:06,120 --> 00:02:10,960 So for example x plus four that's a linear factor and the process is different. 32 00:02:11,010 --> 00:02:15,780 But in this case we have two quadratic vectors and whenever we have a quadratic factor what we want 33 00:02:15,780 --> 00:02:21,650 to put in the numerator is a constant times x plus another constant. 34 00:02:21,650 --> 00:02:28,670 So in this case we'll say X plus B since this fraction over here has a quadratic factor in the denominator. 35 00:02:28,680 --> 00:02:30,570 We wanted to do the same thing. 36 00:02:30,600 --> 00:02:33,160 We've already used constants A and B. 37 00:02:33,180 --> 00:02:40,310 So in this fraction we'll put C X plus D to distinguish the difference between these constants. 38 00:02:40,380 --> 00:02:45,540 Now that we have the set up for our decomposition we want to go ahead and multiply both sides of this 39 00:02:45,540 --> 00:02:48,630 equation by the denominator from the left hand side. 40 00:02:48,840 --> 00:02:53,820 So we're going to multiply both the left and right hand side by the quantity x square plus four times 41 00:02:53,820 --> 00:02:56,400 the quantity x x squared plus 1. 42 00:02:56,520 --> 00:03:00,590 When we do that the entire denominator will cancel from the left hand side. 43 00:03:00,600 --> 00:03:06,430 And of course we'll just be left with x cubed minus 2 x squared plus X plus 1. 44 00:03:06,600 --> 00:03:12,960 When we multiply that denominator by the right hand side specifically this first fraction here the x 45 00:03:12,960 --> 00:03:19,740 squared plus four factor will cancel from the denominator here and we'll just be left with the quantity 46 00:03:19,950 --> 00:03:26,160 x plus being times the quantity x squared plus 1. 47 00:03:26,160 --> 00:03:30,870 The trick I would use for remembering this when I was taking tests is just I'm always left with a factor 48 00:03:30,930 --> 00:03:32,540 that's not in the denominator. 49 00:03:32,540 --> 00:03:39,030 So because X squared plus fours in the denominator here obviously I know that the other factor X squared 50 00:03:39,030 --> 00:03:43,500 plus 1 is the one that's going to be multiplied here by X plus B. 51 00:03:43,620 --> 00:03:48,110 Same thing over here when I multiply this entire denominator by this fraction. 52 00:03:48,150 --> 00:03:55,160 The x squared plus 1 will cancel and I'll be left with the quantity C X plus three times the quantity. 53 00:03:55,170 --> 00:03:59,680 The other factor X squared plus four. 54 00:03:59,820 --> 00:04:05,910 Now that I've eliminated all of my fractions what I want to go ahead and do is expand the right hand 55 00:04:05,910 --> 00:04:06,990 side over here. 56 00:04:06,990 --> 00:04:09,540 In other words multiply all these terms together. 57 00:04:09,570 --> 00:04:16,920 So when I multiply the quantity x plus B times the quantity x squared plus 1 I'll get a x cubed plus 58 00:04:17,050 --> 00:04:22,710 X plus b x squared plus B. 59 00:04:22,710 --> 00:04:29,370 And when I multiply the quantity C X plus the times the quantity x square plus 4 I'll get c x cubed 60 00:04:29,760 --> 00:04:42,760 plus for C X plus D x squared plus for me my goal now here is going to be to collect like terms so I 61 00:04:42,760 --> 00:04:48,310 want to pull together all of my terms involving x cubed all of my terms involving x squared all of my 62 00:04:48,310 --> 00:04:51,560 terms involving X and all of my constants. 63 00:04:51,610 --> 00:04:57,660 So notice here that I have a x cubed and I have C x cubed. 64 00:04:57,730 --> 00:05:03,890 So when I want to do is pull those two terms together and factor out the x cubed. 65 00:05:04,090 --> 00:05:09,480 When I do that I'll get a plus C times x cubed. 66 00:05:09,640 --> 00:05:12,670 I'm going to do the same thing for each of these other factors. 67 00:05:12,850 --> 00:05:17,190 So notice here that I have the x squared and DX squared. 68 00:05:17,230 --> 00:05:25,240 If I pull those together and factor out an x squared I'm going to get plus the quantity B plus D times 69 00:05:25,240 --> 00:05:26,380 x squared. 70 00:05:26,980 --> 00:05:38,020 When I bring mine a x and my 4C x terms together and I factor out an x I'll get plus the quantity a 71 00:05:38,020 --> 00:05:41,600 plus for C times x. 72 00:05:41,800 --> 00:05:48,430 And then when I bring my constants together b and for Dean I don't have to factor out anything because 73 00:05:48,430 --> 00:05:49,690 these are just constant. 74 00:05:49,690 --> 00:05:54,800 So I'll say plus the quantity B plus for D. 75 00:05:54,850 --> 00:06:01,600 The reason for collecting terms like this and factoring out the X variables is now I can equate coefficients 76 00:06:01,600 --> 00:06:04,000 on the left hand on the left and right hand side. 77 00:06:04,300 --> 00:06:07,980 So I can say the coefficient here on x cubed. 78 00:06:07,990 --> 00:06:12,060 Right now this is an empty box because the coefficient is just one. 79 00:06:12,220 --> 00:06:18,190 But I can say that the coefficient on the x cubed term here 1 is equal to the coefficient on the x cubed 80 00:06:18,280 --> 00:06:19,380 term on the right. 81 00:06:19,470 --> 00:06:20,100 A plus. 82 00:06:20,120 --> 00:06:20,800 See. 83 00:06:20,980 --> 00:06:28,340 So I can set those equal to each other and say one is equal to a plus C in the same way. 84 00:06:28,480 --> 00:06:34,900 I can set the convention on the left hand side on the x squared term negative two equal to the coefficient 85 00:06:34,900 --> 00:06:38,080 on the x squared term on the right hand side which is the posti. 86 00:06:38,080 --> 00:06:44,920 So I can say negative 2 is equal to B plus D. 87 00:06:44,920 --> 00:06:52,320 Again I can take the coefficient on the x term here which is just one and set that equal to a plus for 88 00:06:52,320 --> 00:06:55,310 C the coefficient on the x term on the right hand side. 89 00:06:55,330 --> 00:07:00,720 So say one is equal to a plus four scene. 90 00:07:01,150 --> 00:07:07,390 And then finally the constant on the left hand side one can be set equal to the constant on the right 91 00:07:07,390 --> 00:07:10,510 hand side over here B plus four. 92 00:07:10,610 --> 00:07:17,340 So I can say one is equal to B plus four D. 93 00:07:17,350 --> 00:07:22,960 This is basically now a system of simultaneous equations that I can use to solve for the constants A 94 00:07:22,960 --> 00:07:24,790 B C and D. 95 00:07:25,060 --> 00:07:27,930 I've collected my equations in this way. 96 00:07:27,970 --> 00:07:33,520 I've grouped the equations that have only a and c involved here on the left and the equations that only 97 00:07:33,520 --> 00:07:35,030 have B and D involved. 98 00:07:35,080 --> 00:07:38,750 Over here on the right that makes it easy for me to solve for the constant. 99 00:07:38,740 --> 00:07:44,920 So what I'll do over here on the left is subtract this entire first second equation here from the top 100 00:07:44,920 --> 00:07:46,670 equation like this. 101 00:07:46,730 --> 00:07:51,710 When I do that I'll get one minus one which will give me 0 on the left hand side. 102 00:07:52,030 --> 00:07:59,440 I'll get a minus a which again will give me 0 and I'll get C minus for C which will give me a negative 103 00:07:59,530 --> 00:08:07,240 3 C that gives me 0 equals negative 3 C which gives me C is equal to zero. 104 00:08:07,390 --> 00:08:16,040 When I plug that back in to my equation at the top here I'll get one equals a plus zero. 105 00:08:16,210 --> 00:08:19,620 And I can see that one is equal to a. 106 00:08:19,900 --> 00:08:26,520 So now I can start an documenting that I have solved for some of my constants. 107 00:08:26,520 --> 00:08:29,480 I'm going to do the same thing over here to solve for B and D. 108 00:08:29,840 --> 00:08:33,860 I'll subtract this entire second equation from the first equation. 109 00:08:33,870 --> 00:08:38,760 The reason I can do it is because I know looking at this equation that if I subtract the second one 110 00:08:38,760 --> 00:08:43,950 from the first one then my b's will cancel and I'm looking for one of the variables to disappear so 111 00:08:43,950 --> 00:08:45,150 I can solve the other one. 112 00:08:45,270 --> 00:08:47,710 I did the same thing over here with the A's and the C's. 113 00:08:47,730 --> 00:08:51,330 I knew that the A's would cancel when I said a minus a. 114 00:08:51,420 --> 00:08:57,100 So I didn't have to manipulate or do anything extra to this so I subtract that negative to minus 1 which 115 00:08:57,100 --> 00:09:05,310 should give me a negative 3 B minus B will just give me a 0 and D minus 4 D will give me a negative 116 00:09:05,370 --> 00:09:09,580 3 D when I divide both sides by negative 3. 117 00:09:09,590 --> 00:09:12,620 I can see that D is equal to 1. 118 00:09:12,900 --> 00:09:20,280 And when I plug the equals one back into this first equation over here I get negative two equals B plus 119 00:09:20,340 --> 00:09:21,150 1. 120 00:09:21,150 --> 00:09:26,140 Subtract one from both sides I get negative three equals B. 121 00:09:26,370 --> 00:09:32,430 So I can indicate that I've now also solved for B and D. 122 00:09:32,460 --> 00:09:37,290 What this means now that I have all four constants is that I can plug each of these four constants back 123 00:09:37,290 --> 00:09:42,510 into my partial fractions decomposition here and then integrate that. 124 00:09:42,540 --> 00:09:45,970 So I saved our constants up here in the upper right hand corner. 125 00:09:46,140 --> 00:09:49,690 What I'm going to do now is plug them back into our decomposition. 126 00:09:49,690 --> 00:09:54,710 So I'm going to have the integral of here a x plus B over the quantity x where plus 4. 127 00:09:54,870 --> 00:09:57,180 Well we know that a is 1. 128 00:09:57,480 --> 00:10:01,190 So we'll get 1 x and we know that B is negative 3. 129 00:10:01,280 --> 00:10:06,770 So we get plus negative 3 divided by the quantity x square plus 4. 130 00:10:07,110 --> 00:10:10,940 And then we have plus C X plus D when we know that C is zero. 131 00:10:11,160 --> 00:10:20,040 So we get 0 x plus D is 1 and will divide that either quantity x squared plus 1 D X.. 132 00:10:20,040 --> 00:10:32,310 Now I want to simplify this so I'll get X minus 3 over x squared plus four plus one over the quantity 133 00:10:32,310 --> 00:10:36,960 x squared plus 1 D X with any partial fractions. 134 00:10:36,960 --> 00:10:41,190 The composition problem once you get to this point you're probably going to want to look at breaking 135 00:10:41,190 --> 00:10:46,960 these fractions apart as much as possible into separate integrals to try to make this easier. 136 00:10:46,980 --> 00:10:52,830 So what we'll do is we'll break apart this first fraction X minus three divided by x squared plus four 137 00:10:52,890 --> 00:11:00,260 into two and we'll break them into separate integrals so we'll get the integral of x squared plus for 138 00:11:01,060 --> 00:11:11,060 the X minus three times the integral of one over x squared plus for D x. 139 00:11:11,070 --> 00:11:14,850 So what we did there is we separated it into two fractions. 140 00:11:14,850 --> 00:11:18,320 This first term in the numerator X divided by x squared plus four. 141 00:11:18,570 --> 00:11:22,920 And the second term in the numerator negative three divided by x squared plus four. 142 00:11:23,070 --> 00:11:28,830 And in the process we pulled the negative three out in front of the integral to leave just one over 143 00:11:28,890 --> 00:11:30,810 x squared plus four. 144 00:11:30,870 --> 00:11:38,970 We can't simplifier a second fraction at all so we'll still just say plus one over the quantity x squared 145 00:11:39,240 --> 00:11:47,040 plus 1 the X and now that we have this simplified we want to go ahead and start integrating the first 146 00:11:47,040 --> 00:11:50,130 fraction here X divided by x squared plus four. 147 00:11:50,160 --> 00:11:57,640 We're going to need use substitution for will set you equal to x squared plus four. 148 00:11:57,930 --> 00:12:06,810 We'll take the derivative to get D-New which will be 2 x times d x and then we'll solve for the X to 149 00:12:06,810 --> 00:12:13,140 get the x is equal to d u over to X. 150 00:12:13,170 --> 00:12:18,840 Our second and third integrals will be using this formula up here that I didn't mention yet but the 151 00:12:18,930 --> 00:12:24,750 formula is for the integral of one over x squared plus any constant. 152 00:12:24,750 --> 00:12:28,610 So in this case with our second integral here a will be two. 153 00:12:28,620 --> 00:12:33,650 Because notice we're taking a constant Square is a constant and we're saying a squared. 154 00:12:33,660 --> 00:12:38,230 So we're looking for essentially the square root of the second term here. 155 00:12:38,340 --> 00:12:40,640 And the square root of four is two. 156 00:12:40,800 --> 00:12:43,920 So we know that a in this formula will be 2. 157 00:12:43,920 --> 00:12:51,900 And what this formula tells us is that 1 divided by x squared plus a constant squared is equal to 1 158 00:12:51,900 --> 00:12:55,690 divided by that constant times tangent to the negative one. 159 00:12:55,690 --> 00:12:59,420 Or the inverse tangent of X divided by that constant again. 160 00:12:59,610 --> 00:13:02,640 And of course C the constant of integration. 161 00:13:02,700 --> 00:13:07,970 So we're going to be using that formula for the second and third integrals that we have here. 162 00:13:08,010 --> 00:13:13,060 So let's go ahead and for this first integral make our substitutions with you. 163 00:13:13,160 --> 00:13:19,740 So we'll say X over you because remember we said X equal to the denominator x square plus 4. 164 00:13:20,040 --> 00:13:26,980 And then we solved for the X so for dx we got you over to X.. 165 00:13:27,470 --> 00:13:31,420 So those two are multiplied together and then we have our second and third integrals here. 166 00:13:31,430 --> 00:13:33,390 So we're minus three. 167 00:13:33,560 --> 00:13:37,550 Now the integral of 1 divided by x squared plus four. 168 00:13:37,580 --> 00:13:43,340 Again we use this formula and notice that we're setting everything equal to this here so we'll get 1 169 00:13:43,400 --> 00:13:45,110 divided by a. 170 00:13:45,110 --> 00:13:50,200 We know that a is going to be 2 because we're taking the square root of this. 171 00:13:50,300 --> 00:13:55,310 The constants in this position here and square root here of four is two. 172 00:13:55,570 --> 00:14:04,220 So we'll get two times tangent to the negative one of x divided by again a as two we'll leave the constant 173 00:14:04,220 --> 00:14:06,890 immigration for the end of our integration. 174 00:14:06,890 --> 00:14:14,400 So now we take the integral of the third integral here and we'll get 1 divided by 1. 175 00:14:14,450 --> 00:14:21,320 Because the constant here is just one the square root of which is 1 so 1 over 1 tangent to the negative 176 00:14:21,320 --> 00:14:24,600 one of x divided by 1. 177 00:14:24,770 --> 00:14:28,470 And now we can add our plus c our constant integration. 178 00:14:28,670 --> 00:14:31,270 We want to go ahead and simplify as much as we can. 179 00:14:31,280 --> 00:14:38,690 Notice that we'll get the X here to cancel and we can pull the one half out in front so we'll get 1 180 00:14:38,690 --> 00:14:49,440 1/2 times the integral of one over you do you minus we'll say three halves tangent to the negative one 181 00:14:49,860 --> 00:15:00,030 of X over two plus tangent to the negative one of x plus C. 182 00:15:00,380 --> 00:15:06,290 And when we integrate one over you remember that the integral of one to over a variable like this is 183 00:15:06,290 --> 00:15:08,920 the natural log of the absolute value of that variable. 184 00:15:09,140 --> 00:15:14,120 So we'll get 1 1/2 times the natural log of the absolute value of you. 185 00:15:14,660 --> 00:15:18,920 And then the rest of our integration 186 00:15:24,570 --> 00:15:27,750 our final step is just to plug back in for you. 187 00:15:27,960 --> 00:15:31,290 So we'll get one halftimes the natural log of the absolute value. 188 00:15:31,300 --> 00:15:33,930 Remember that U is x squared plus 4. 189 00:15:34,170 --> 00:15:46,070 So x squared plus four and then minus three halves tangent to the negative one of x over two plus tangent 190 00:15:46,140 --> 00:15:52,430 the negative one of x and you can go ahead and leave it in this form sometimes you'll factor out of 191 00:15:52,430 --> 00:15:57,260 one half just to make this a little bit simpler and get rid of the coefficient I the natural log in 192 00:15:57,260 --> 00:15:59,100 the Tench the negative one here. 193 00:15:59,100 --> 00:16:00,410 But you don't need to. 194 00:16:00,410 --> 00:16:03,330 You can leave it in this form and this is your final answer. 21606