# Homework Help: Integration of powers of ln x

1. May 26, 2008

### ritwik06

1. The problem statement, all variables and given/known data

Find
$$\int ln x^{2009}$$
2. Relevant equations

The integration by parts!

3. 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

2. May 26, 2008

### ritwik06

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

3. May 26, 2008

### Vid

4. May 26, 2008

### ritwik06

Hey, have u found out the pattern urself?

5. May 26, 2008

### DavidWhitbeck

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!

6. May 26, 2008

### Defennder

7. May 27, 2008

### 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.

8. May 27, 2008

### 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.

9. May 27, 2008

### 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.

10. May 27, 2008

### 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)