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Integral of ln x. Created by Sal Khan.
Video transcript
The goal of this video is to try to figure out the anti derivative of the natural log of x. And it's not completely obvious how to approach this at first, even if I was to tell you to use integration by parts. You'd say integration by parts? You're looking for the anti derivative of something that can be expressed as a product of two functions. Looks like I only have one function right over here, the natural log of x. But it might become a little bit more obvious if I were to rewrite this. As the integral of the natural log of x times one dx. Now you do have the product of two functions. One is a function. A function of x. It's not actually dependent on x, it's actually going to be one, but you can have fx equal to one. And now it might become a little more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as as a product of one function and the derivative of another function and the derivative of another function, and this is really just the reverse product rule. We've shown that multiple times all ready. This is going to be equal to the product of both functions. F of x times g of x times g of x minus, minus the anti-derivative of, instead of having f in g-prime, you're gonna f-prime in g. So f prime of x, f prime of x times g of x, g of x, dx, dx. And we've seen this multiple times. So when you figure out what, what should be f and what should be g, for f, you want to figure out something that it's easy to take the derivative of and it simplifies things, possibly, if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So a good candidate for f of x is natural log of x. If you were to take the derivative of it, it's one over x. Let me write this down. So let's say that f of x is equal to the natural log of x, of x. And f prime of x is equal to one over x. And let's set g Prime of x is equal to one. So g Prime of x is equal to one. That means that g of x could be equal to, could be equal to x. And so, let's go back right over here. So this is going to be equal to. This is going to be equal to f(x) times g(x). Well, f(x) times g(x), is x natural log of x. So, g(x) is x and f(x) is the natural log of x. I just like writing the x in front of the natural log, to avoid ambiguity. So this is x natural log of x, minus the antiderivative of f prime of x, which is one over x, times g of x, which is x, which is x, d x, d x. Well what's this going to be equal to? Well what we have inside, the integrant, this is just one over x times x. Which is just equal to one. So this simplifies quite nicely. This is going to end up equalling, this is going to end up equalling, I can go, let me put it right there. This is going to end up equalling, alright, x natural log of x, natural log of x minus the anti-derivative of just dx, or the anti-derivative of 1dx, or the integral of 1dx. I should say, or the anti-derivative of one, is just minus x, and this is just an anti-derivative of this; if we want to write the entire class of anti-derivatives, we just have to add, we just have to add a plus c here. And we are done. We figured out the antiderivative of the natural log of x. And I'd encourage you to take the derivative of this. For this part you're gonna use the product rule, and verify that you do indeed get natural log of x when you take the derivative of this.