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Integration of powers of ln x

  • Thread starter ritwik06
  • Start date
  • #1
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Homework Statement



Find
[tex]\int ln x^{2009}[/tex]

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) [tex]ln x^{2}[/tex]=x ([tex]ln x^{2}[/tex])-2x ln x +2x
3)[tex]ln x^{3}[/tex]=x[tex]ln x^{3}[/tex]-3x[tex]ln x^{2}[/tex]+6x ln x-6x
4)[tex]ln x^{4}[/tex]=x[tex]ln x^{4}[/tex]-4x[tex]ln x^{3}[/tex]+12x [tex]ln x^{2}[/tex]-12x ln x +12x


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

Answers and Replies

  • #2
580
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Please note that all the powers here are not on x but on (ln x)
 
  • #3
Vid
401
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  • #4
580
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Hey, have u found out the pattern urself?
 
  • #5
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 [itex]I_n = \int (\ln x)^n dx[/itex] must satisfy and from that you can figure out the series. So keep working, it's not too bad!
 
  • #7
Vid
401
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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.
 
  • #8
Defennder
Homework Helper
2,591
5
Here's what I have:

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

And so on...

Actually I think it's the same thing, only with different variables.
 
  • #9
Vid
401
0
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.
 
  • #10
63
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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)
 

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