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Integration by parts: ∫ln(x)dx

This video shows how to find the antiderivative of the natural log of x using integration by parts. We rewrite the integral as ln(x) times 1dx, then choose f(x) = ln(x) and g'(x) = 1. The antiderivative is xln(x) - x + C. Created by Sal Khan.

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Video transcript

The goal of this video is to try to figure out the antiderivative of the natural log of x. And it's not completely obvious how to approach this at first, even if I were to tell you to use integration by parts, you'll say, integration by parts, you're looking for the antiderivative of something that can be expressed as the product of two functions. It 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 1dx. 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 always going to be 1, but you could have f of x is equal to 1. And now it might become a little bit more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as the product of one function, and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already. This is going to be equal to the product of both functions, f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times. So when you figure out 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 good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. So g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x, I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x times g of x, which is x, which is xdx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely. This is going to end up equaling x natural log of x minus the antiderivative of, just dx, or the antiderivative of 1dx, or the integral of 1dx, or the antiderivative of 1 is just minus x. And this is just an antiderivative of this. If we want to write the entire class of antiderivatives we just have to add a plus c here, and we are done. We figured out the antiderivative of the natural log of x. I encourage you to take the derivative of this. For this part, you're going to use the product rule and verify that you do indeed get natural log of x when you take the derivative of this.