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Current time:0:00Total duration:8:40

𝘶-substitution: challenging application

Video transcript

I was just looking on the discussion boards on the Khan Academy Facebook page and but Denny put up this problem asking for it to be solved and it seems like a problem of general interest it's the indefinite integral of two to the natural log of x over everything over X DX and on the message board a beacon also put up a solution and it is the correct solution but I thought this was of general interest so I'll I'll make a quick video on it so the first thing when you see an integral like this is you say hey you know I have this natural log of X up in the in the numerator and you know what where do I start and the first thing that should maybe pop out of you is that this is the same thing as the integral of 1 over X times 2 to the natural log of X DX and so you have an expression here or it's kind of part of our larger function and you have its derivative right we know that the derivative let me write over here we know that the derivative with respect to X of the natural log of X is equal to 1 over X so we have some expression and we have its derivative which tells us that we can use u-substitution sometimes you can do it in your head but this problem it's still not trivial to do it in your head so let's make the substitution let's substitute this right here with a u so let's do that so if you define u and it doesn't have to be you it's just that's the convention it's called u substitution it could have been s substitution for all we care but let's say U is equal to the natural log of X and then D u DX the derivative of U with respect to X of course is equal to 1 over X or just the differential D u D U if we just multiply both sides by DX is equal to 1 over X DX so let's make our substitution this is our integral so this will be equal to the indefinite integral or the antiderivative of 2 to the now u so 2 to the U times 1 over X DX now what is 1 over X DX that's just D u so this term times that term is just our D U let me do it in a different color one over x times DX is just equal to D U that's just equal to that thing right there now this still doesn't look like an easy integral although it's gotten simplified a good bit and to solve that you know whenever I see the my my it be the variable that I'm integrating against in the exponent you know we don't have any easy exponent rules here the only time that the only thing that I'm familiar with where I have my X or my my variable that I'm integrating against in my exponent is the case of e to the X we know that the integral of e to the X DX is equal to e to the X plus C so if I could somehow turn this into some variation of e to the X maybe or e to the U maybe I can make this integral a little bit more tractable so let's see how can we redefine how can we redefine this right here well 2 2 is equal to what 2 is the same thing as e to the natural log of 2 right the natural log of 2 is the power you have to raise 2 to get you have to raise e to to get 2 so if you raise e to that power you're of course going to get two this is actually the definition of really the natural log you raise e to the natural log two you're going to get two so let's rewrite this using this I guess we could call this this rewrite or this I don't to call it quite a substitution it's just a different way of writing the number two so this will be equal to this will be equal to instead of writing the number two I could write e to the natural log of two and all of that to the you d u and now what is this equal to well well if I take something to an exponent and then to another exponent this is the same thing as taking my base to the product of those exponents so this is equal to this is let me switch colors this is equal to the integral of e to the u e to the let me write it this way e to the natural log of two e to the natural log of two times you I'm just multiplying these two exponents if I raise something to something then raise it again we know from our exponent rules that is just the product of those two exponents d u now this is just a constant factor right here this could be you know this could just be some number and we could use a calculator to figure out what this is you know we could set this equal to a but we know in general that the integral this is pretty straightforward we've now put it in this form the antiderivative of e to the a you D U is just 1 over a e to the a u this comes from this definition up here and of course plus C and the chain rule if we take the derivative of this we take the derivative of the inside which is just going to be a we multiply that times the 1 over a it cancels out and we're just left with E to the au so this definitely works out so the antiderivative of this thing right here is going to be equal to it's going to be equal to 1 over our a it's going to be one of our constant term 1 over the natural log of 2 times our whole expression e E and I'm going to do something I'm just going to you know I could write this is just some number times U so I can write it as u times some number and I'm just doing that to put it in a form that might help us simplify it a little bit so it's e to the U times the natural log of 2 right all I did is I swapped this order I could have written this as e to the natural log of 2 times u I'm just you know this is an a a times U is the same thing as u times a plus C so this is our answer but we have to kind of reverse substitute before we can feel satisfied that we've taken the antiderivative with respect to X but before I do that let's see if I can simplify this a little bit what is if I have just from our natural log properties or our logarithm a times the natural log of B we know this is the same thing as the natural log of B to the a let me draw a line here right that this becomes the exponent on whatever we're taking the natural log of so you let me write this here u times the natural log of 2 is the same thing as the natural log of 2 to the U so we can rewrite our antiderivative as being equal to 1 over the natural log of 2 that's just that part here times e to the this can be rewritten based on this logarithm property as the natural log of 2 to the U and of course we still have our plus C there now what is e raised to the natural log of 2 to the U the natural log of 2 to the U is the power that you have to raise e to to get to 2 to the U right by definition so if we raise e to that power what are we going to get we're going to get 2 to the U so this is going to be equal to 1 over the natural log of 2 this simplifies to just 2 to the U I drew it up here that the natural log a I could just write in general terms let me do it up here and maybe I'm beating a dead horse but I can in general write any number a is being equal to e to the natural log of a this is the exponent you have to raise e to to get a if you raise e to that you're going to get a so e to the natural log of 2 to the U that's just 2 to the U and then I have my plus C of course and now we can reverse substitute what did we set u equal to we defined you up here is equal to the natural log of X so let's just reverse substitute right here and so the answer to our original equation your answer to let me write it here because it's satisfying when you see it to this kind of fairly fairly convoluted looking antiderivative problem 2 to the natural log of x over X DX we now find is equal to we just replace U with natural log of x because that was our substitution 1 over the natural log of 2 times 2 to the natural log of X plus C and we're done this isn't in the denominator the way I wrote it might look a little ambiguous and we're done and that was a pretty neat problem so thanks thanks to bud for for posting that