How to Integrate ln(x) Using Integration by Parts | Proving the Solution

[MysticDude]

Homework Statement

Okay so today I started to learn about integration by parts and I understand the basics of it and can do some of the simpler problems, but this one made me stop.
I have to integrate ln(x).
I know the answer is xln(x) - x but I have to prove it.

Homework Equations

\int u dv = uv - \int v du

The Attempt at a Solution

Okay so what I do is make a table and list my u,du,v,dv. Here is my table:
u = ln(x); v = 1(because ln(x) = 1*ln(x))
du = 1/x; dv = 0;

Okay so the start is ln(x) instead of xln(x) so that's not a good first step. And the \int vdu part is equal to 1(x(\frac{1}{x}) = 1). At this part I'm stuck. I mean I could change the table for the start to be xln(x) but that doesn't make sense to me because where the x come from? I checked Wolfram Alpha but the steps confused me for this problem.

I'll appreciate any help!

:D

 

[fss]

Okay so what I do is make a table and list my u,du,v,dv. Here is my table:
u = ln(x); v = 1(because ln(x) = 1*ln(x))
du = 1/x; dv = 0;

Reconsider the part bolded above. If u = ln x, you still have something left in that integral to account for...

 

[MysticDude]

Reconsider the part bolded above. If u = ln x, you still have something left in that integral to account for...

OH Shize! I forgot about the dx! But wait, if I use v = dx, then what will dv equal to? Or am I getting this wrong and dv = dx and v = x?

 

[Dick]

Use u=ln(x) and v=x. So, sure, dv=dx. v=1 doesn't make much sense.

 

[MysticDude]

Use u=ln(x) and v=x. v=1 doesn't make much sense.

Okay question: do I use v = x because the deriv of x is dx? I mean since v = x then dv = dx. That's the only way that I could think of using x. Am I right? I understand how to use the equation, it was just getting the u and the v that was confusing me a lot.

 

[Dick]

Okay question: do I use v = x because the deriv of x is dx? I mean since v = x then dv = dx. That's the only way that I could think of using x. Am I right? I understand how to use the equation, it was just getting the u and the v that was confusing me a lot.

Well, sure. If u=ln(x) and v=x then u*dv=ln(x)*dx. That's the integral you want to solve on the left side, yes? So what's the right side?

 

[MysticDude]

Well since the equation(on the right side) is uv - \int vdu I would get xln(x) - x. I get the -x part because x(1/x) = 1, so I'm integrating 1, which is x. I could also go ahead and factor out the x making it x(ln(x) -1). I know that this is the answer but I needed to get v, thanks Dick and fss. Much appreciated!

 

[danielatha4]

n integrations of ln(x):

x^n[ln(x)-(1/n!)(S)]

where S is the sigma sum of 1/k where k=1 to n

 

[MysticDude]

n integrations of ln(x):

x^n[ln(x)-(1/n!)(S)]

where S is the sigma sum of 1/k where k=1 to n

So you are saying that if I was to take the...let's say, 5th integral of ln(x) I would do x^{5}[ln(x)- \frac{1}{5!}(\sum_{k=1}^{5}\frac{1}{k})]? That's pretty cool! Just even by typing in 1 I would get xln(x) - x, that's awesome! Gonna show my teacher this!