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These are the user uploaded subtitles that are being translated: 1 00:00:00,460 --> 00:00:05,190 In this video we're doing another partial fractions problem but we're talking about distinct quadratic 2 00:00:05,190 --> 00:00:06,070 factors. 3 00:00:06,090 --> 00:00:11,700 So we've been given this problem the integral of x square plus X plus 1 divided by quantity to X plus 4 00:00:11,700 --> 00:00:18,060 one times quantity x squared plus 1 and we need to use partial fractions to evaluate this integral So 5 00:00:18,060 --> 00:00:22,670 remember with partial fractions what we want to do is come up with a partial fractions decomposition 6 00:00:22,680 --> 00:00:28,170 we want to simplify this fraction into its partial fractions decomposition so that we can integrate 7 00:00:28,170 --> 00:00:33,480 that decomposition instead of this fraction which we can't integrate as is. 8 00:00:33,480 --> 00:00:35,350 So how are we going to use partial fractions. 9 00:00:35,370 --> 00:00:40,170 Well what we need to do first is recognize that we have distinct factors because we have a factor of 10 00:00:40,170 --> 00:00:43,080 2 x plus 1 and a factor of x squared plus 1. 11 00:00:43,080 --> 00:00:45,780 So they are distinct they're different from each other. 12 00:00:46,080 --> 00:00:51,660 And then we have to recognize that 2 x plus 1 is a linear factor because this is X to the first power 13 00:00:52,200 --> 00:00:56,750 and an X squared plus one is a quadratic factor because we have an x squared term. 14 00:00:56,760 --> 00:00:58,800 So here's what our decompositions going to look like. 15 00:00:58,800 --> 00:01:03,000 First of all we're going to take the original fraction we're going to put that on the left hand side 16 00:01:03,000 --> 00:01:05,890 of this new equation that we're going to create. 17 00:01:05,910 --> 00:01:11,120 So we're going to have 2 x plus one times quantity x squared plus 1. 18 00:01:11,190 --> 00:01:13,680 OK so the original fraction exactly as is. 19 00:01:13,850 --> 00:01:19,460 But on the right hand side when we have a linear factor we just put a single constant in the numerator 20 00:01:19,470 --> 00:01:23,030 so we're going to do a divided by 2 x plus 1. 21 00:01:23,100 --> 00:01:25,530 The single constant over the linear factor. 22 00:01:25,710 --> 00:01:30,600 But then we're going to add to that here's where we put the quadratic factor X squared plus 1. 23 00:01:30,810 --> 00:01:36,430 But whenever we have a quadratic factor we have to say B X plus C. 24 00:01:36,450 --> 00:01:39,430 So we started with B because A was already taken. 25 00:01:39,480 --> 00:01:45,420 If we had another linear factor here we would say plus D divided by the linear factor. 26 00:01:45,420 --> 00:01:53,370 And if we had another quadratic factor we would say e x plus F divided by the other quadratic factor. 27 00:01:53,370 --> 00:01:59,650 So that gives you an idea of how to set up a decomposition with linear factors and quadratic factors. 28 00:01:59,790 --> 00:02:05,760 So now that we have this decomposition we need to simplify it and what we'll do is we'll multiply both 29 00:02:05,760 --> 00:02:14,820 sides of this equation as always by the denominator from the left hand side so to X plus 1 times x squared 30 00:02:14,820 --> 00:02:16,080 plus 1. 31 00:02:16,080 --> 00:02:19,660 When we do that will multiply both of these factors by the left hand side. 32 00:02:19,770 --> 00:02:24,450 We're going to get this entire denominator to cancel because this 2 x plus 1 will cancel with this 2 33 00:02:24,450 --> 00:02:28,940 x plus 1 this x squared plus 1 will cancel with this x squared plus 1. 34 00:02:29,010 --> 00:02:34,840 Leaving us with just x squared plus X plus 1 the numerator from the left hand side. 35 00:02:34,980 --> 00:02:41,400 Then over here we multiply both of these factors by a divided by 2 x plus 1 will get the 2 x plus 1 36 00:02:41,400 --> 00:02:43,750 to cancel with this 2 x plus 1. 37 00:02:43,890 --> 00:02:47,840 Leaving us with just a times x squared plus 1. 38 00:02:47,850 --> 00:02:54,780 The other factor then is here when we multiply 2 x plus 1 Times Square plus 1 by this b x plus C divided 39 00:02:54,780 --> 00:02:59,910 by x squared plus 1 will get this x squared plus 1 to cancel with this x squared plus 1. 40 00:02:59,940 --> 00:03:07,800 Leaving us with just the factor of 2 x plus 1 so will get plus B X plus C is important parentheses around 41 00:03:07,800 --> 00:03:11,430 that multiplied by 2 x plus 1. 42 00:03:11,430 --> 00:03:16,640 Now what we want to do is simplify the right hand side which will do by multiplying everything out. 43 00:03:16,650 --> 00:03:23,010 So we'll distribute this across the x square plus 1 and we'll get a x squared plus a will foil out the 44 00:03:23,010 --> 00:03:31,640 b x plus C times quantity 2 x plus 1 and will get X times 2 x is to be x squared. 45 00:03:31,760 --> 00:03:40,290 We'll take B X times 1 and get B X. we'll take C times 2 x and get to see X and we'll take C times 1 46 00:03:40,320 --> 00:03:41,890 and get c. 47 00:03:41,910 --> 00:03:45,710 Now what we want to do is collect like terms. 48 00:03:45,750 --> 00:03:53,640 So we're going to take all of our x squared terms together we'll say a x squared plus two B x squared. 49 00:03:53,640 --> 00:03:55,760 That takes care of these two. 50 00:03:55,830 --> 00:04:00,980 Then we'll take all of our x terms together so we'll say let's go ahead and put parentheses around these. 51 00:04:01,070 --> 00:04:08,210 So are x terms we're going to have B X plus to see X that takes care of these two and then all grouped 52 00:04:08,210 --> 00:04:14,810 together our constants or constants are going to be a plus C taking care of our last two terms then 53 00:04:14,820 --> 00:04:22,020 what we want to do is factor out the x variable from each of these sets Apprentice's here so here we're 54 00:04:22,020 --> 00:04:23,750 going to factor out an x squared. 55 00:04:23,760 --> 00:04:30,690 So we're going to say a plus to B will be the only thing left when we factor in x squared out of X squared 56 00:04:30,690 --> 00:04:32,310 plus 2 be x squared. 57 00:04:32,310 --> 00:04:34,060 We pulled out the x squared. 58 00:04:34,140 --> 00:04:39,360 We were just left with the A and A to B here in the second set of parentheses we're going to pull out 59 00:04:39,360 --> 00:04:40,120 an x. 60 00:04:40,260 --> 00:04:48,300 Leaving us with only B plus 2 c multiplied by X and then we'll leave our constants as they are. 61 00:04:48,300 --> 00:04:53,730 So now what we want to do is recognize that over here on the left hand side we have 1 x squared plus 62 00:04:53,730 --> 00:04:56,040 1 x plus 1. 63 00:04:56,070 --> 00:04:59,780 What we want to do is equate coefficients from the left and the right hand side. 64 00:04:59,960 --> 00:05:05,840 So we can say that the coefficient on x squared on the left hand side is one the coefficient on x square 65 00:05:05,840 --> 00:05:13,580 on the right hand side is a plus to B we can say that the coefficient on X is 1 and the coefficient 66 00:05:13,580 --> 00:05:16,410 on X is B plus 2 c. 67 00:05:16,730 --> 00:05:23,180 And then finally that over here the constant is 1 and the constant over here is a plus C. 68 00:05:23,210 --> 00:05:28,670 So now with that in mind what we can do is we can write equations where we set these things equal to 69 00:05:28,670 --> 00:05:29,500 one another. 70 00:05:29,510 --> 00:05:40,310 So we'll go ahead and say a plus two B is equal to 1 being plus 2 C is equal to 1 and a plus C is equal 71 00:05:40,310 --> 00:05:41,480 to 1. 72 00:05:41,480 --> 00:05:45,310 Now with this in mind let's go ahead and eliminate one of the variables. 73 00:05:45,320 --> 00:05:46,370 We have three variables. 74 00:05:46,370 --> 00:05:47,870 Let's go ahead and eliminate a. 75 00:05:47,870 --> 00:05:51,790 So this last equation here a plus C is equal to 1. 76 00:05:51,950 --> 00:05:59,300 We'll go ahead and subtract C from both sides and let's say a is equal to 1 minus C since a is equal 77 00:05:59,300 --> 00:06:03,820 to one minus C we can plug one minus C in for a into this first equation. 78 00:06:03,950 --> 00:06:10,210 So instead of a rate here we're going to get 1 minus C plus 2 B is equal to 1. 79 00:06:10,250 --> 00:06:13,250 If we rewrite this we'll keep that to be. 80 00:06:13,250 --> 00:06:19,760 We keep our minus C's all say to B minus C and then subtract 1 from both sides and we'll get to B minus 81 00:06:19,760 --> 00:06:21,400 C is equal to zero. 82 00:06:21,650 --> 00:06:28,580 Then we'll take our second equation here and put it with it so we'll say B plus 2 C is equal to 1. 83 00:06:28,580 --> 00:06:32,990 Now what I want to do here is multiply through this second equation the left and the right hand side 84 00:06:32,990 --> 00:06:35,160 all the terms by 2. 85 00:06:35,270 --> 00:06:40,250 The reason is because if I multiply everything by two I'm going to get to B and then I can subtract 86 00:06:40,250 --> 00:06:41,480 one equation from the other. 87 00:06:41,510 --> 00:06:44,300 I'll get 2 B minus 2 B which is zero. 88 00:06:44,300 --> 00:06:48,480 In other words I'll get my B's to cancel so I'll be able to solve for C. 89 00:06:48,650 --> 00:06:55,210 So if I multiply through by 2 instead of B I'll get to B instead of two c I'll get 4 C. 90 00:06:55,280 --> 00:06:58,760 And instead of one here I'll get two. 91 00:06:58,760 --> 00:07:06,020 Now what I can do is subtract this bottom equation from the top equation and the result is going to 92 00:07:06,020 --> 00:07:13,130 be to B minus two B which is zero so those cancel negative C minus 4 C's going to give me a negative 93 00:07:13,220 --> 00:07:17,710 5 C and zero minus 2 is going to give me a negative 2. 94 00:07:17,990 --> 00:07:23,570 If I divide both sides by negative 5 I get C is equal to negative 2 divided by negative 5. 95 00:07:23,660 --> 00:07:26,990 The negative signs cancel and I get a positive two fifths. 96 00:07:26,990 --> 00:07:30,870 Now I need to use this value of c to find values for a and b. 97 00:07:30,920 --> 00:07:34,660 I already know a is equal to 1 minus C since C is two fifths. 98 00:07:34,670 --> 00:07:38,780 I can say is equal to 1 minus two fifths. 99 00:07:38,870 --> 00:07:44,070 I can call 1 here instead of 1 or call it 5. 100 00:07:44,190 --> 00:07:51,170 So I have a common denominator and I'll get a is equal to 5 managed to is three or three fifths. 101 00:07:51,170 --> 00:07:58,370 Now let's use this second equation here here again this one to find a value for B will go out and subtract 102 00:07:58,400 --> 00:08:04,970 to see from both sides and we'll get B is equal to 1 minus 2 c plugging in two fifths for C. 103 00:08:04,970 --> 00:08:11,480 I'm going to get B is equal to 1 minus two times two fifths is four fifths. 104 00:08:11,480 --> 00:08:12,830 So I'll get four fifths. 105 00:08:13,060 --> 00:08:19,760 I'll change my one to five deaths that I can have a common denominator and I'll get B is equal to 5 106 00:08:19,760 --> 00:08:22,770 months for as 1 so I get 1 fifth. 107 00:08:22,820 --> 00:08:25,430 So now I know that A is three fifths. 108 00:08:25,430 --> 00:08:28,540 B is one fifth and C is two fifths. 109 00:08:28,610 --> 00:08:34,670 With that in mind I can plug into my partial fraction decomposition and instead of a here I put three 110 00:08:34,670 --> 00:08:37,060 fifths instead of B. 111 00:08:37,100 --> 00:08:42,210 I'll put one fifth and instead of C I'll put two fifths. 112 00:08:42,320 --> 00:08:48,920 And now this value here this entire right hand side is my partial fraction decomposition with the values 113 00:08:48,920 --> 00:08:50,760 that I found for a b and c.. 114 00:08:50,990 --> 00:08:56,870 This is what I'm going to use to replace the original fraction and this will be a much easier integral 115 00:08:57,080 --> 00:08:58,580 to evaluate. 116 00:08:58,580 --> 00:09:03,800 So now instead of the original integral I can say the integral of and take this whole value here and 117 00:09:03,800 --> 00:09:14,420 find that so I'm going to say three fifths divided by 2 x plus one plus one fifth X plus two fifths 118 00:09:14,960 --> 00:09:19,170 divided by x squared plus 1 D x. 119 00:09:19,220 --> 00:09:22,070 Now we can work on simplifying this integral. 120 00:09:22,070 --> 00:09:25,520 So what we're going to do is we're going to split it into two separate integrals and we're going to 121 00:09:25,520 --> 00:09:26,960 factor out the constants. 122 00:09:27,170 --> 00:09:31,130 So we're going to bring this three fifths out in front and we're going to say three fifths times the 123 00:09:31,130 --> 00:09:39,080 integral of one divided by 2 x plus 1 D X then I'm going to say plus here I'm going to factor out a 124 00:09:39,080 --> 00:09:44,470 wonderful song and say one fifth time's the integral of x plus two. 125 00:09:44,690 --> 00:09:49,230 Because when I multiply two by one fifth I get two fifths this value right here. 126 00:09:49,430 --> 00:09:55,490 So I have to say X plus two divided by x squared plus 1 D x. 127 00:09:55,490 --> 00:10:00,960 Now before I start evaluating integrals I also want to split the second integral into two integrals 128 00:10:00,960 --> 00:10:01,560 of it's own. 129 00:10:01,560 --> 00:10:06,750 So whenever you have multiple terms in the numerator of a fraction here you can split them into separate 130 00:10:06,750 --> 00:10:12,540 fractions so instead of X plus two divided by x squared plus one I can say X divided by x squared plus 131 00:10:12,540 --> 00:10:17,400 one plus and a separate fraction to divide it by x squared plus 1. 132 00:10:17,400 --> 00:10:25,530 So if I rewrite this I'm going to say three fifths times the integral one over to X plus 1 D x plus 133 00:10:25,620 --> 00:10:33,080 one fifth times the integral of x divided by x squared plus 1 x plus. 134 00:10:33,090 --> 00:10:35,520 And I have to keep my one fifth here. 135 00:10:35,670 --> 00:10:44,330 So plus one fifth time's the integral of two divided by x squared plus 1 the X and this first integral. 136 00:10:44,340 --> 00:10:49,830 I can easily evaluate using my formula here remember that the integral of 1 divided by X is going to 137 00:10:49,830 --> 00:10:52,440 be the natural log of the absolute value of x. 138 00:10:52,440 --> 00:10:56,970 In other words whatever is in the denominator there is going to go inside of my absolute value bars 139 00:10:57,000 --> 00:10:59,300 in the argument of my natural log function. 140 00:10:59,430 --> 00:11:04,500 So I'm just going to take this three fifths leave it there as a coefficients of three fifths and then 141 00:11:04,500 --> 00:11:10,870 I'm going to take my denominator to X plus 1 and say natural log of the absolute value of 2 x plus 1. 142 00:11:11,070 --> 00:11:16,230 And the only thing I have to be careful of is I always have to apply changeroom and divide it by the 143 00:11:16,230 --> 00:11:18,210 derivative of the denominator. 144 00:11:18,210 --> 00:11:22,560 So here if you just have x by itself the derivative of X is 1. 145 00:11:22,740 --> 00:11:24,720 So you would divide this by 1. 146 00:11:24,720 --> 00:11:27,610 But of course that doesn't change anything which is why you don't see it. 147 00:11:27,900 --> 00:11:29,760 But here we have 2 x plus 1. 148 00:11:29,790 --> 00:11:32,370 The derivative of 2 x plus 1 is 2. 149 00:11:32,370 --> 00:11:38,540 So I have to divide this by two which is the same as multiplying by 1 1/2. 150 00:11:38,550 --> 00:11:40,540 So that takes care of our first integral. 151 00:11:40,830 --> 00:11:43,230 The third integral here the last integral. 152 00:11:43,230 --> 00:11:48,810 Keep in mind that I can pull this two out in front so I can call this one and change this to a 2. 153 00:11:48,820 --> 00:11:50,680 Now I have two fifths out in front. 154 00:11:51,000 --> 00:11:55,900 We know that the integral of 1 divided by x squared plus 1 that's a very common function. 155 00:11:56,100 --> 00:11:57,810 That's the inverse tangent function. 156 00:11:57,810 --> 00:12:05,150 So I can say plus two fifths times arctan or the inverse tangent function of x. 157 00:12:05,160 --> 00:12:10,980 When you have the exact form 1 divided by x squared plus 1 the integral of that is the inverse tangent 158 00:12:10,980 --> 00:12:12,690 function of x. 159 00:12:12,720 --> 00:12:17,610 So that takes care of our last integral there and then this middle integral here will evaluate using 160 00:12:17,610 --> 00:12:18,600 u substitution. 161 00:12:18,600 --> 00:12:25,440 So for now let's just go ahead and say plus one fifth times the integral of x divided by x squared plus 162 00:12:25,440 --> 00:12:27,810 1 D x. 163 00:12:27,810 --> 00:12:35,160 Now for that integral we'll say you is equal to x squared plus 1 the derivative of u d u is going to 164 00:12:35,160 --> 00:12:42,300 be equal to 2 x and then we add the D X and then if we solve this for Dayaks we get x is equal to D 165 00:12:42,300 --> 00:12:44,460 you divide it by 2 x. 166 00:12:44,460 --> 00:12:49,920 So now if we go ahead and make our substitutions for that integral we have that coefficient of 1 fifth 167 00:12:50,340 --> 00:12:57,840 times the integral will keep that x in the numerator so x divided by we know X squared plus 1 is you. 168 00:12:57,850 --> 00:13:06,690 So we'll have you in the denominator and X is going to be D u over to X which we just found here and 169 00:13:06,690 --> 00:13:10,850 then we'll go ahead and add our other two terms. 170 00:13:10,890 --> 00:13:15,000 Now inside of the integral you can see we're going to get X and the numerator to cancel with X in the 171 00:13:15,000 --> 00:13:16,030 denominator. 172 00:13:16,290 --> 00:13:18,950 This to here we can pull out in front. 173 00:13:18,960 --> 00:13:20,710 This is a 2 in the denominator. 174 00:13:20,790 --> 00:13:23,600 So it comes and moves into the denominator here. 175 00:13:23,760 --> 00:13:30,460 Multiply it by the five and we end up with one tenth times the integral of one over you. 176 00:13:30,610 --> 00:13:37,170 D'you hear we're going to multiply this one half by three fifths and we're going to get plus three over 177 00:13:37,290 --> 00:13:47,550 10 natural log absoute value of 2 x plus 1 plus two fifths arctan or inverse tangent function of x and 178 00:13:47,550 --> 00:13:54,330 then here for our integral the integral of one over you will be natural log of the absolute value of 179 00:13:54,330 --> 00:14:02,620 you and then we'll go ahead and add to that the other terms we know that you as equal to x squared plus 180 00:14:02,620 --> 00:14:03,060 one. 181 00:14:03,070 --> 00:14:10,030 So we'll say one tenth natural log of the absolute value of x squared plus one will add these other 182 00:14:10,030 --> 00:14:19,360 terms so three tenths natural log of absolute value to X plus one plus two fifths times inverse tangent 183 00:14:19,360 --> 00:14:27,130 function of x and then we have to add the constant c to account for that constant of integration. 184 00:14:27,130 --> 00:14:29,510 Now you could leave your final answer just like this. 185 00:14:29,530 --> 00:14:36,100 Or if you wanted to you could factor out a one tenth and if you wanted to factor out one tenth you would 186 00:14:36,100 --> 00:14:38,840 pull the one tent out in front like this. 187 00:14:39,040 --> 00:14:41,320 And then we need to make sure we pulled out of every turn. 188 00:14:41,380 --> 00:14:48,010 So obviously it's going to come out of this natural log absolute value of x squared plus 1. 189 00:14:48,100 --> 00:14:52,840 Then we're also going to pull it out of three over 10 the 10 is going to get pulled out here from the 190 00:14:52,840 --> 00:14:53,770 denominator. 191 00:14:53,770 --> 00:14:56,000 But the three and the numerator is going to remain. 192 00:14:56,020 --> 00:15:01,500 So we have plus three natural log absolute value to X plus 1. 193 00:15:01,780 --> 00:15:04,780 And then here pulling one tenth out of two fifths. 194 00:15:04,780 --> 00:15:06,510 So how do we do that easily. 195 00:15:06,730 --> 00:15:10,840 Well basically all we're saying is we're going to factor a one tenth out of the two fifths. 196 00:15:10,840 --> 00:15:15,580 So what do we have to multiply by one tenth in order to get two fifths. 197 00:15:15,610 --> 00:15:18,860 And if you don't know how to do that in your head here's what you do. 198 00:15:18,880 --> 00:15:24,620 What do we have to multiply by one tenth to get two fifths. 199 00:15:24,760 --> 00:15:30,790 And then you would just multiply both sides by 10 and you would get x is equal to 20 over 5 for x is 200 00:15:30,790 --> 00:15:32,770 equal to 4. 201 00:15:32,800 --> 00:15:40,930 So when we factor 110 out of two fifths what's left over is force we would say plus for inverse tangent 202 00:15:41,470 --> 00:15:47,560 of X and then we would close our brackets and then say plus C and that should make sense to us this 203 00:15:47,660 --> 00:15:53,990 4 because if we multiply four by one tenth we get four over 10 which is the same as two over five. 204 00:15:54,190 --> 00:16:00,080 So this is your final answer with the one Tende factored out in front in order to simplify it. 23029