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### Course: Integral Calculus>Unit 1

Lesson 17: Integration by parts

# 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.

## Want to join the conversation?

• f(x) = 1/ln(x),

is there a vertical asymptote at x=0 ?

I think that there is...is there?

Also, is the point x=1 a global minima?
I think not because ln(1) = 0 and therefore f(x) is not define there so it can't be minima.

What do you think?

Thanks.
(8 votes)
• I don't think there is a vertical asymptote at x=0 because lim(x-->0+) f(x) = 0
(3 votes)
• When you differentiate the end result, don't you get ln(x)-1 rather than ln(x)?
(12 votes)
• The calculation follows the chain rule : d/dx (x ln x ) = 1 * ln x + x * 1/x = ln x + 1
So, in d/dx (x ln x - x) you have to add d/dx (-x) = -1
Together : = ln x + 1 - 1 = ln x
(4 votes)
• Can xlnx be written as ln x^2 ?
If not then why ?
(0 votes)
• If you are trying to use properties of logarithms, you would bring the "x" from the front into the logarithm as an exponent, resulting in:

ln (x^x)
(39 votes)
• At sal say the word integrand. What is the difference between an integral and an integrand?
(10 votes)
• An integral is the whole operator: ∫ f(x) dx
An integrand is just the function you are integrating. So for ∫ 3x^2 dx, the integrand is 3x^2.
(16 votes)
• At he integrated g'(x)=1 to get g(x)=x, but shouldn't the integral of g'(x)=1 be g(x)=x+c?
(8 votes)
• That is correct if that is where you were going to end your problem, but since there will be further integrals down the road, you can just add a +C at the end of the problem to encompass all the +C you would have had to put in.
(5 votes)
• why do you consider 1 as a function in this case and not in other cases?
(6 votes)
• You are going to see more and more of this if you continue in math, that is, the creative use of the rules and properties of numbers and processes. In this case, treating the 1 as the result of differentiating some function g(x)=x, made it possible the use of integration by parts to solve the problem. Use whatever works to solve problems. Get creative. But stay within the rules. For me, this is the most fun part of math where you can unleash your creativity! At its best, it is the playground of new ideas, at its worst, it is where you hone your intuition by learning what works and what doesn't - and that isn't bad at all!
(8 votes)
• Hi, just doing some revision for this, I always thought that the integral of Ln(x) was always 1/x?
(2 votes)
• No, the derivative of ln(x) is 1/x. As Sal points out here, ∫ lnx dx is
xlnx-x+c
(10 votes)
• How do I know which part of the function is f(x) and which is g'(x)? I always end up trying both possibilities.
(3 votes)
• You need to develop an intuition for which function will simplify with either taking the derivative or the anti-derivative. It's a matter of practice.
(5 votes)
• I am 65 years old, and re-learning calculus on my own, using an old textbook.
When I first come upon integrating ln x, the book has not yet mentioned integration by parts. How would you integrate ln x without using integration by parts?
(5 votes)
• Awesome! I'm in my mid 50's.

Your book probably just provided the cookbook result. I'm not aware of any other method to compute the integral other than IBP.
(2 votes)
• If the derivitive of a function is written as the prime of that function (i.e. f'(x)) is there such notation for the antiderivitive of a function?
(2 votes)

## 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.