How can substitution help with the integration of powers of ln x?

[ritwik06]

Homework Statement

Find
\int ln x^{2009}

Homework Equations

The integration by parts!

The Attempt at a Solution

I have been trying to discover a series which might have helped:
I have taken out the integrals of :
1) ln x = x ln x -x
2) ln x^{2}=x (ln x^{2})-2x ln x +2x
3)ln x^{3}=xln x^{3}-3xln x^{2}+6x ln x-6x
4)ln x^{4}=xln x^{4}-4xln x^{3}+12x ln x^{2}-12x ln x +12x


But I am unable to find a series. Please help!

 

[ritwik06]

Please note that all the powers here are not on x but on (ln x)

 

[Vid]

Do a u sub with u = ln(x). Then, do integration by parts.
To make the by parts quicker do it with tabular to get the pattern then use series notation.
http://en.wikipedia.org/wiki/Integration_by_parts#Tabular_integration_by_parts

 

[ritwik06]

Hey, have u found out the pattern urself?

 

[DavidWhitbeck]

Hey, have u found out the pattern urself?

I got it as a series doing it just as Vid said. By integrating by parts you can find a recurrence relation that I_n = \int (\ln x)^n dx must satisfy and from that you can figure out the series. So keep working, it's not too bad!

 

[Defennder]

Do a u sub with u = ln(x). Then, do integration by parts.
To make the by parts quicker do it with tabular to get the pattern then use series notation.
http://en.wikipedia.org/wiki/Integration_by_parts#Tabular_integration_by_parts

I think you meant it should be a substitution with u = (lnx)^2009.

 

[Vid]

Maybe I'm missing something, but that sub got me Int( (ln(x))^2009)dx = 1/2009Int(e^(u^(1/2009))*u^(1/2009)du. The first sub got me Int(u^(2009)*e^(u))du. The second one is much easier to get to and to solve.

 

[Defennder]

Here's what I have:

u = (ln x)^{2009} \ du=2009\frac{{lnx}^{2008}}{x}
dv=1 \ v = x
\int (lnx)^{2009} dx = x(lnx)^{2009} - 2009\int (lnx)^{2008} dx

And so on...

Actually I think it's the same thing, only with different variables.

 

[Vid]

Yea, my u was for a change of variable followed by a by parts. Yours was just a by parts so it's the same. I had thought you meant to do a change of variable with u = ln(x)^2009 which is pretty messy.

 

[Raze2dust]

or maybe you can substitute u=ln(x), which would leave you with integrals in terms of e^(u) which, i think, will be easier to work with as compared to ln(x)