Homework Help: Need help with integration by parts

1. Dec 4, 2012

banshee43

Hi all this is my first post hopefully i do it right.
1. The problem statement, all variables and given/known data
integrate ln^2(6x)dx

3. The attempt at a solution
*integral* ln^2(6x)dx
u=ln^2(6x) dv=dx
du=(2ln(6x))/x dx v=x

xln^2(6x)-*integral*x(2ln(6x))/x dx

xln^2(6x)-2*integral*ln(6x) dx

u=ln(6x) dv=dx
du=1/x dx v=x

xln^2(6x)-2(xln(6x)-*integral*x(1/x)dx)
xln^2(6x)-2(xln(6x)-*integral*dx)

MY ANSWER.... that is not correct

xln^2(6x)-2xln(6x)-2x+Constant

i do not know where i am going wrong and i think im using parts correctly..i dont see any place where substitution could be used but i could be wrong

2. Dec 4, 2012

Dick

Hah. Well done! I think the only problem is that the -2x should be +2x. You should be able to find where that mistake happened pretty easily.

3. Dec 4, 2012

banshee43

is it because the -2 is distributed and not +2?

4. Dec 4, 2012

Dick

If you mean what I think, yes. xln^2(6x)-2(xln(6x)-*integral*dx). (-2)*(-1)=+2.

5. Dec 4, 2012

banshee43

yes! thank you so much.... those simple mistakes will be the death of me!

6. Dec 5, 2012

SteamKing

Staff Emeritus
In your u substitution, have you applied the chain rule correctly?

7. Dec 5, 2012

banshee43

Yes, I believe have.
Using prime notation:

ln^2(6x)'

u=ln^2(6x)
u'=2ln(6x)ln(6x)'
u'=2ln(6x)/(6x)*(6x)'
u'=2ln(6x)/(6x)*6
since ((6x^1)' = *Const*x^n=*Const*nx^n-1 in my case 6x^1 = 1*6x^1-1
the 6's cancel and you are left with
u'=2ln(6x)/x :)