# Integration of powers of ln x

## 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)$$ln x^{3}$$=x$$ln x^{3}$$-3x$$ln x^{2}$$+6x ln x-6x
4)$$ln x^{4}$$=x$$ln x^{4}$$-4x$$ln x^{3}$$+12x $$ln x^{2}$$-12x ln x +12x

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

Vid
Hey, have u found out the pattern urself?

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
Homework Helper
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
Homework Helper
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.

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)