1 00:00:00,000 --> 00:00:00,550 2 00:00:00,550 --> 00:00:03,790 I was just looking on the discussion boards on the Khan 3 00:00:03,790 --> 00:00:08,150 Academy Facebook page, and Bud Denny put up this problem, 4 00:00:08,150 --> 00:00:09,335 asking for it to be solved. 5 00:00:09,335 --> 00:00:12,010 And it seems like a problem of general interest. 6 00:00:12,009 --> 00:00:18,509 If the indefinite integral of 2 to the natural log of x 7 00:00:18,510 --> 00:00:23,200 over, everything over x, dx. 8 00:00:23,199 --> 00:00:27,210 And on the message board, Abhi Khanna also put up a solution, 9 00:00:27,210 --> 00:00:29,109 and it is the correct solution, but I thought this was of 10 00:00:29,109 --> 00:00:32,079 general interest, so I'll make a quick video on it. 11 00:00:32,079 --> 00:00:34,420 So the first thing when you see an integral like this, is you 12 00:00:34,420 --> 00:00:36,635 say, hey, you know, I have this natural log of x up in 13 00:00:36,634 --> 00:00:40,149 the numerator, and where do I start? 14 00:00:40,149 --> 00:00:43,149 And the first thing that should maybe pop out at you, is that 15 00:00:43,149 --> 00:00:49,369 this is the same thing as the integral of one over x times 2 16 00:00:49,369 --> 00:00:53,869 to the natural log of x, dx. 17 00:00:53,869 --> 00:00:57,579 And so you have an expression here, or it's kind of part of 18 00:00:57,579 --> 00:01:00,640 our larger function, and you have its derivative, right? 19 00:01:00,640 --> 00:01:03,570 We know that the derivative, let me write it over here, we 20 00:01:03,570 --> 00:01:06,260 know that the derivative with respect to x of the natural 21 00:01:06,260 --> 00:01:09,870 log of x is equal to 1/x. 22 00:01:09,870 --> 00:01:13,870 So we have some expression, and we have its derivative, which 23 00:01:13,870 --> 00:01:15,980 tells us that we can use substitution. 24 00:01:15,980 --> 00:01:18,250 Sometimes you can do in your head, but this problem, it's 25 00:01:18,250 --> 00:01:19,689 still not trivial to do in your head. 26 00:01:19,689 --> 00:01:21,129 So let's make the substitution. 27 00:01:21,129 --> 00:01:24,569 Let's substitute this right here with a u. 28 00:01:24,569 --> 00:01:26,709 So let's do that. 29 00:01:26,709 --> 00:01:29,890 So if you define u, and it doesn't have to be u, it's 30 00:01:29,890 --> 00:01:32,670 just, that's the convention, it's called u-substitution, it 31 00:01:32,670 --> 00:01:35,120 could have been s-substitution for all we care. 32 00:01:35,120 --> 00:01:39,760 Let's say u is equal to the natural log of x, and then du 33 00:01:39,760 --> 00:01:43,400 dx, the derivative of u with respect to x, of course 34 00:01:43,400 --> 00:01:45,219 is equal to 1/x. 35 00:01:45,219 --> 00:01:49,480 Or, just the differential du, if we just multiply both sides 36 00:01:49,480 --> 00:01:53,600 by dx is equal to 1 over x dx. 37 00:01:53,599 --> 00:01:55,079 So let's make our substitution. 38 00:01:55,079 --> 00:01:55,989 This is our integral. 39 00:01:55,989 --> 00:02:00,159 So this will be equal to the indefinite integral, or the 40 00:02:00,159 --> 00:02:05,479 antiderivative, of 2 to the now u, so 2 to the u, 41 00:02:05,480 --> 00:02:08,450 times 1 over x dx. 42 00:02:08,449 --> 00:02:10,619 Now what is 1 over x dx? 43 00:02:10,620 --> 00:02:11,439 That's just du. 44 00:02:11,439 --> 00:02:13,859 So this term times that term is just our du. 45 00:02:13,860 --> 00:02:15,540 Let me do it in a different color. 46 00:02:15,539 --> 00:02:20,379 1 over x times dx is just equal to du. 47 00:02:20,379 --> 00:02:23,530 That's just equal to that thing, right there. 48 00:02:23,530 --> 00:02:26,629 Now, this still doesn't look like an easy integral, although 49 00:02:26,629 --> 00:02:29,229 it's gotten simplified a good bit. 50 00:02:29,229 --> 00:02:34,000 And to solve, you know, whenever I see the variable 51 00:02:34,000 --> 00:02:36,710 that I'm integrating against in the exponent, you know, 52 00:02:36,710 --> 00:02:38,570 we don't have any easy exponent rules here. 53 00:02:38,569 --> 00:02:42,709 The only thing that I'm familiar with, where I have my 54 00:02:42,710 --> 00:02:45,909 x or my variable that I'm integrating against in 55 00:02:45,909 --> 00:02:48,430 my exponent, is the case of e to the x. 56 00:02:48,430 --> 00:02:51,990 We know that the integral of e to the x, dx, is equal 57 00:02:51,990 --> 00:02:55,090 to e to the x plus c. 58 00:02:55,090 --> 00:02:58,469 So if I could somehow turn this into some variation of e to the 59 00:02:58,469 --> 00:03:02,370 x, maybe, or e to the u, maybe I can make this integral a 60 00:03:02,370 --> 00:03:03,980 little bit more tractable. 61 00:03:03,979 --> 00:03:04,599 So let's see. 62 00:03:04,599 --> 00:03:10,889 How can we redefine this right here? 63 00:03:10,889 --> 00:03:14,379 Well, 2, 2 is equal to what? 64 00:03:14,379 --> 00:03:19,969 2 is the same thing as e to the natural log of 2, right? 65 00:03:19,969 --> 00:03:22,849 The natural log of 2 is the power you have to 66 00:03:22,849 --> 00:03:26,859 raise e to to get 2. 67 00:03:26,860 --> 00:03:28,620 So if you raise e to that power, you're, of 68 00:03:28,620 --> 00:03:29,409 course going to get 2. 69 00:03:29,409 --> 00:03:32,030 This is actually the definition of really, the natural log. 70 00:03:32,030 --> 00:03:35,490 You raise e to the natural log of 2, you're going to get 2. 71 00:03:35,490 --> 00:03:38,180 So let's rewrite this, using this-- I guess we could call 72 00:03:38,180 --> 00:03:40,650 this this rewrite or-- I don't want to call it 73 00:03:40,650 --> 00:03:41,379 quite a substitution. 74 00:03:41,379 --> 00:03:44,439 It's just a different way of writing the number 2. 75 00:03:44,439 --> 00:03:48,939 So this will be equal to, instead of writing the number 76 00:03:48,939 --> 00:03:53,259 2, I could write e to the natural log of 2. 77 00:03:53,259 --> 00:03:58,090 And all of that to the u du. 78 00:03:58,090 --> 00:04:00,120 And now what is this equal to? 79 00:04:00,120 --> 00:04:02,689 Well, if I take something to an exponent, and then to another 80 00:04:02,689 --> 00:04:05,419 exponent, this is the same thing as taking my base to the 81 00:04:05,419 --> 00:04:07,250 product of those exponents. 82 00:04:07,250 --> 00:04:12,810 So this is equal to, let me switch colors, this is equal 83 00:04:12,810 --> 00:04:21,620 to the integral of e, to the u, e to the, let 84 00:04:21,620 --> 00:04:22,639 me write it this way. 85 00:04:22,639 --> 00:04:27,659 e to the natural log of 2 times u. 86 00:04:27,660 --> 00:04:29,770 I'm just multiplying these two exponents. 87 00:04:29,769 --> 00:04:32,189 I raise something to something, then raise it again, we know 88 00:04:32,189 --> 00:04:34,170 from our exponent rules, it's just a product of 89 00:04:34,170 --> 00:04:35,500 those two exponents. 90 00:04:35,500 --> 00:04:37,250 du. 91 00:04:37,250 --> 00:04:39,810 Now, this is just a constant factor, right here. 92 00:04:39,810 --> 00:04:42,209 This could be, you know, this could just be some number. 93 00:04:42,209 --> 00:04:44,299 We could use a calculator to figure out what this is. 94 00:04:44,300 --> 00:04:45,720 We could set this equal to a. 95 00:04:45,720 --> 00:04:48,050 But we know in general that the integral, this is pretty 96 00:04:48,050 --> 00:04:50,710 straightforward, we've now put it in this form. 97 00:04:50,709 --> 00:04:56,289 The antiderivative of e to the au, du, is just 98 00:04:56,290 --> 00:04:59,750 1 over a e to the au. 99 00:04:59,750 --> 00:05:02,519 This comes from this definition up here, and of course plus 100 00:05:02,519 --> 00:05:03,779 c, and the chain rule. 101 00:05:03,779 --> 00:05:05,974 If we take the derivative of this, we take the derivative 102 00:05:05,975 --> 00:05:08,210 of the inside, which is just going to be a. 103 00:05:08,209 --> 00:05:10,509 We multiply that times the one over a, it cancels 104 00:05:10,509 --> 00:05:12,920 out, and we're just left with e to the au. 105 00:05:12,920 --> 00:05:14,420 So this definitely works out. 106 00:05:14,420 --> 00:05:18,590 So the antiderivative of this thing right here is going to be 107 00:05:18,589 --> 00:05:24,339 equal to 1 over our a, it's going to be 1 over our constant 108 00:05:24,339 --> 00:05:29,639 term, 1 over the natural log of 2 times our whole 109 00:05:29,639 --> 00:05:32,485 expression, e e. 110 00:05:32,485 --> 00:05:35,030 And I'm going to do something. 111 00:05:35,029 --> 00:05:38,169 This is just some number times u, so I can write it as 112 00:05:38,170 --> 00:05:39,250 u times some number. 113 00:05:39,250 --> 00:05:42,649 And I'm just doing that to put in a form that might help us 114 00:05:42,649 --> 00:05:44,049 simplify it a little bit. 115 00:05:44,050 --> 00:05:47,270 So it's e to the u times the natural log of 2, right? 116 00:05:47,269 --> 00:05:48,779 All I did, is I swapped this order. 117 00:05:48,779 --> 00:05:50,639 I could have written this as e to the natural 118 00:05:50,639 --> 00:05:53,139 log of 2 times u. 119 00:05:53,139 --> 00:05:56,589 If this an a, a times u is the same thing as u times a. 120 00:05:56,589 --> 00:05:58,529 Plus c. 121 00:05:58,529 --> 00:06:01,109 So this is our answer, but we have to kind of reverse 122 00:06:01,110 --> 00:06:04,340 substitute before we can feel satisfied that we've taken 123 00:06:04,339 --> 00:06:07,299 the antiderivative with respect to x. 124 00:06:07,300 --> 00:06:09,680 But before I do that, let's see if I can simplify 125 00:06:09,680 --> 00:06:10,879 this a little bit. 126 00:06:10,879 --> 00:06:15,550 What is, if I have, just from our natural log properties, 127 00:06:15,550 --> 00:06:18,449 or logarithms, a times the natural log of b. 128 00:06:18,449 --> 00:06:20,139 We know this is the same thing as the natural 129 00:06:20,139 --> 00:06:21,479 log of b to the a. 130 00:06:21,480 --> 00:06:22,900 Let me draw a line here. 131 00:06:22,899 --> 00:06:23,219 Right? 132 00:06:23,220 --> 00:06:25,130 That this becomes the exponent on whatever we're taking 133 00:06:25,129 --> 00:06:26,620 the natural log of. 134 00:06:26,620 --> 00:06:31,949 So u, let me write this here, u times the natural log of 135 00:06:31,949 --> 00:06:35,469 2, is the same thing as the natural log of 2 to the u. 136 00:06:35,470 --> 00:06:38,440 So we can rewrite our antiderivative as being equal 137 00:06:38,439 --> 00:06:42,019 to 1 over the natural log of 2, that's just that part here, 138 00:06:42,019 --> 00:06:46,449 times e to the, this can be rewritten based on this 139 00:06:46,449 --> 00:06:51,740 logarithm property, as the natural log of 2 to the u, and 140 00:06:51,740 --> 00:06:55,750 of course we still have our plus c there. 141 00:06:55,750 --> 00:07:00,500 Now, what is e raised to the natural log of 2 to the u? 142 00:07:00,500 --> 00:07:03,259 The natural log of 2 to the u is the power that you have to 143 00:07:03,259 --> 00:07:06,959 raise e to to get to 2 to the u, right? 144 00:07:06,959 --> 00:07:08,039 By definition! 145 00:07:08,040 --> 00:07:11,540 So if we raise e to that power, what are we going to get? 146 00:07:11,540 --> 00:07:13,040 We're going to get 2 to the u. 147 00:07:13,040 --> 00:07:17,080 So this is going to be equal to 1 over the natural log of 2. 148 00:07:17,079 --> 00:07:20,349 This simplifies to just 2 to the u. 149 00:07:20,350 --> 00:07:21,220 I drew it up here. 150 00:07:21,220 --> 00:07:24,970 The natural log a I could just write in general terms, let 151 00:07:24,970 --> 00:07:27,520 me do it up here, and maybe I'm beating a dead horse. 152 00:07:27,519 --> 00:07:30,569 But I can in general write any number a as being equal to 153 00:07:30,569 --> 00:07:32,889 e to the natural log of a. 154 00:07:32,889 --> 00:07:36,229 This is the exponent you have to raise e to to get a. 155 00:07:36,230 --> 00:07:38,879 If you raise e to that, you're going to get a. 156 00:07:38,879 --> 00:07:42,420 So e to the natural log of 2 to the u, that's just 2 to the u. 157 00:07:42,420 --> 00:07:44,900 And then I have my plus c, of course. 158 00:07:44,899 --> 00:07:46,449 And now we can reverse substitute. 159 00:07:46,449 --> 00:07:48,829 What did we set u equal to? 160 00:07:48,829 --> 00:07:53,060 We defined u, up here, as equal to the natural log of x. 161 00:07:53,060 --> 00:07:55,990 So let's just reverse substitute right here. 162 00:07:55,990 --> 00:08:00,569 And so the answer to our original equation, your answer 163 00:08:00,569 --> 00:08:03,259 to, let me write it here, because it's satisfying when 164 00:08:03,259 --> 00:08:08,949 you see it, to this kind of fairly convoluted-looking 165 00:08:08,949 --> 00:08:14,729 antiderivative problem, 2 to the natural log of x over x dx, 166 00:08:14,730 --> 00:08:17,814 we now find is equal to, we just replaced u with natural 167 00:08:17,814 --> 00:08:21,750 log of x, because that was our substitution, and 1 over the 168 00:08:21,750 --> 00:08:29,410 natural log of 2 times 2 to the natural log of x plus c. 169 00:08:29,410 --> 00:08:30,100 And we're done. 170 00:08:30,100 --> 00:08:31,970 This isn't in the denominator, the way I wrote it might 171 00:08:31,970 --> 00:08:33,410 look a little ambiguous. 172 00:08:33,409 --> 00:08:34,110 And we're done! 173 00:08:34,110 --> 00:08:37,340 And that was a pretty neat problem, and so thanks 174 00:08:37,340 --> 00:08:39,899 to Bud for posting that.