(2^(ln x))/x antiderivative example Finding ∫(2^ln x)/x dx
(2^(ln x))/x antiderivative example
- 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.
- 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.
- 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.
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