Integration by parts: ∫ln(x)dx (video) | Khan Academy

Video transcript

The finish of this video recording is to try to figure out the antiderivative of the natural log of x. And it ‘s not wholly obvious how to approach this at beginning, 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 logarithm of x. But it might become a little snatch 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 affair of ten. It ‘s not actually dependent on ten, it ‘s always going to be 1, but you could have f of ten is equal to 1. And now it might become a small 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 officiate, and the derivative of another function, and this is actually 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, farad of x times g of x minus the antiderivative of, alternatively of having f and gigabyte prime, you ‘re going to have f prime and g. So degree fahrenheit prime of x times guanine of x dx. And we ‘ve seen this multiple times. therefore 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 derived function of and it simplifies things, possibly if you ‘re taking the derivative of it. And for gigabyte premier of x, you want to find something where it ‘s easily to take the antiderivative of it. so good campaigner for f of ten 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 ten is equal to the natural log of x. then farad flower of x is equal to 1 over ten. And let ‘s set g prime of x is peer to 1. so g prime of x is equal to 1. That means that deoxyguanosine monophosphate of ten could be peer to x. And indeed let ‘s go back properly over here. So this is going to be peer to f of x times thousand of x. Well, degree fahrenheit of x times guanine of x is adam natural log of x. So gigabyte of ten is ten, and f of x is the natural log of x, I just like writing the x in battlefront of the natural log of ten to avoid ambiguity. So this is ten natural log of x minus the antiderivative of degree fahrenheit 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 merely 1 over ten times x, which is merely equal to 1. so this simplifies quite nicely. This is going to end up equaling x natural log of x minus the antiderivative of, merely dx, or the antiderivative of 1dx, or the integral of 1dx, or the antiderivative of 1 is equitable minus x. And this is just an antiderivative of this. If we want to write the entire classify of antiderivatives we precisely have to add a plus hundred here, and we are done. We figured out the antiderivative of the natural log of x. I encourage you to take the derived function of this. For this part, you ‘re going to use the product principle and verify that you do indeed get natural logarithm of ten when you take the derivative of this.

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