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# 𝘶-substitution: challenging application

Finding  ∫(2^ln x)/x dx. Created by Sal Khan.

## Want to join the conversation?

• There is a much easier way to do this problem, that I came up with.
We know that d/dx (a^x) = a^x lna. Then we can figure out that d/dx (a^x/lna) = a^x because lna is a constant and it can be taken out, then it gets cancelled.
So, antiderivative of a^x dx = a^x/lna !
Then we just have to put u = ln x. We get 2^u which is in the form a^x
(13 votes)
• that's also how i would have done this saves a lot of work if you remember that
(1 vote)
• Really dumb question I know, but how come 2^(ln x) = x^(ln 2)
(3 votes)
• Not a dumb question.

The key is knowing that 2 = e^ln 2
Review the logarithm properties if you need to on that because that's really useful.

So then
(e^ln 2)^ln x = (e^ln x)^ln 2
exponent properties.

If that part is confusing I like to recall something like
(2^2)^4 = (2^4)^2 = 2^(4*2) = 256

then last we basically repeat the first step
We know e^ln x = x
So
(e^ln x)^ln 2 = x ^(ln 2)

Hope that helps.
(16 votes)
• A silly question, but I dont seem to find the answer... is there any difference between u` and du/dx or is it just a simple matter of choice which notation one uses?
(3 votes)
• That's not a silly question at all. Generally u' and du/dx are simply alternative notations for the same concept, with u' being the notation that originated with Newton and du/dx originating with Leibniz. You need to know both notations because they're used interchangeably.

But there's more to the story. Newton's notation is more compact and often convenient, such as when we state the product rule, (uv)' = uv' + u'v. Leibniz's is more precise because it specifies that the derivative is with respect to the variable x (which is merely assumed in Newton's notation). It's also more powerful because it invites us to think of du and dx as infinitesimal quantities, which is helpful in understanding integration, and provides a bridge to further topics such as multivariable calculus and differential equations. It's been suggested that mathematicians on the European continent outpaced British mathematicians for many decades following the invention of calculus because they used the more flexible du/dx notation on the continent while the British stuck with u' out of loyalty to Newton.
(13 votes)
• Hi, couldn't you use the formula Integral a^x dx = (a^x/ln a) + c
(8 votes)
• Yes thats correct. Sal was just explaining the concept without using that formula.
(3 votes)
• @ I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?
(3 votes)
• e^x and lnx are opposite functions. It would be akin to multiplying by some number and then dividing by that same number. The operations cancel one another out.

e to the natural log of elephant = elephant.
(11 votes)
• how to integrate ln(x)^2
(3 votes)
• you have to use recursive integration by parts(which is product rule in reverse):
⌠ln(x)^2dx = ln(x)•⌠ln(x)dx - ⌠ln(x)/x dx;
⌠ln(x)dx = x•ln(x) - ⌠x/xdx= x•ln(x) - x (+C) //we plug this back in
⌠ln(x)^2dx = ln(x)•(x•ln(x) - x) - ⌠ln(x)/x dx; u=ln(x), u' = 1/x; dx=du/(1/x);
. . . . . . . . =ln(x)•(x•ln(x) - x) - ⌠1/(x/x)du;
. . . . . . . . =ln(x)•(x•ln(x) - x) - u +C
. . . . . . . . =ln(x)•(x•ln(x) - x)- ln(x) +C
. . . . . . . . =ln(x)•(x•ln(x) - x - 1)+C ; hard, but not as bad as (sec(x))^3.
(4 votes)
• You could have taken u=2^ln x differentiated and then substituted right?
(3 votes)
• I'm not sure what you're suggesting, but I suspect it wouldn't work. To make a u-substitution work, the formula has to include both the thing you're choosing as u and the derivative of u (that is, du). Sal is able to do a u-substitution using ln x here because the formula also includes 1/x, the derivative of ln x. We can't do a u-substitution using 2^(ln x) because the formula doesn't contain anything corresponding to the derivative of that expression.
(2 votes)
• I did this myself, and came up with a different answer...
I converted 2 to e^ln(2) first, so the problem would become
INTEGRAL[e^(ln(2)*ln(x))/x dx] I set u = ln(2)*ln(x) and found du to be ln(2)/x.
So 1/ln(2) * INTEGRAL[e^u dx] = 1/ln(2) * e^u
= e^(ln(2)*ln(x))/ln(2) + C

I'd like to know if this is right as well, and if not -- where I went wrong.
Thank you very much :)
(2 votes)
• You have the same answer, you just didn't simplify. Remember that

e^[ log (2) ∙ log (x) ]
= e^[log (2^log(x))]
= 2^log (x)
so once simplified your answer is:
[2^log x ] / log 2 + C

Please be sure you know your log and exponent properties very thoroughly, as you will use them extensively in calculus.
(3 votes)
• after getting ⌠2^u du , can't we use the power rule?

⌠2^u du = (2^(u+1))/(u+1) = 2^(ln(x)+1)/ln(x)+1
(3 votes)
• Can you make a video solving ∫ (x² ) arcsin x dx
(2 votes)

## Video transcript

I was just looking on the discussion boards on the Khan Academy Facebook page, and Bud Denny put up this problem, asking for it to be solved. And it seems like a problem of general interest. If the indefinite integral of 2 to the natural log of x over, everything over x, dx. And on the message board, Abhi Khanna also put up a solution, and it is the correct solution, but I thought this was of general interest, so 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 numerator, and where do I start? And the first thing that should maybe pop out at you, is that this is the same thing as the integral of one 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 it over here, we know that the derivative with respect to x of the natural log of x is equal to 1/x. So we have some expression, and we have its derivative, which tells us that we can use substitution. Sometimes you can do in your head, but this problem, it's still not trivial to do 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 u, it's just, that's the convention, it's called u-substitution, it could have been s-substitution for all we care. Let's say u is equal to the natural log of x, and then du dx, the derivative of u with respect to x, of course is equal to 1/x. Or, just the differential du, 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 du. So this term times that term is just our du. Let me do it in a different color. 1 over x times dx is just equal to du. 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, you know, whenever I see the variable that I'm integrating against in the exponent, you know, we don't have any easy exponent rules here. The only thing that I'm familiar with, where I have my x or 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 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 e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? 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, 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 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. 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 au, du, is just 1 over a e to the au. 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 one 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 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. 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 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. If this 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 logarithms, 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 u, 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. 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 as 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 u, up here, as 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 convoluted-looking antiderivative problem, 2 to the natural log of x over x dx, we now find is equal to, we just replaced u with natural log of x, because that was our substitution, and 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, and so thanks to Bud for posting that.