How Do You Solve the Integral of ln^2(6x) Using Integration by Parts?

Click For Summary

Homework Help Overview

The discussion revolves around the integration of the function ln^2(6x) with respect to x, specifically using integration by parts. Participants are exploring the steps involved in the integration process and examining potential errors in their calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply integration by parts, with one participant expressing uncertainty about their calculations and another questioning the correctness of the chain rule application in their u substitution.

Discussion Status

Some participants have provided feedback on the calculations, noting a possible sign error in the final expression. There is an ongoing examination of the steps taken in the integration process, with no clear consensus reached yet.

Contextual Notes

Participants are discussing the application of integration techniques and the potential for simple mistakes in algebraic manipulation. There is a focus on ensuring that the chain rule is applied correctly during the substitution process.

banshee43
Messages
15
Reaction score
0
Hi all this is my first post hopefully i do it right.

Homework Statement


integrate ln^2(6x)dx

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 I am using parts correctly..i don't see any place where substitution could be used but i could be wrong
 
Physics news on Phys.org
banshee43 said:
Hi all this is my first post hopefully i do it right.

Homework Statement


integrate ln^2(6x)dx



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 I am using parts correctly..i don't see any place where substitution could be used but i could be wrong

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.
 
is it because the -2 is distributed and not +2?
 
banshee43 said:
is it because the -2 is distributed and not +2?

If you mean what I think, yes. xln^2(6x)-2(xln(6x)-*integral*dx). (-2)*(-1)=+2.
 
yes! thank you so much... those simple mistakes will be the death of me!
 
In your u substitution, have you applied the chain rule correctly?
 
SteamKing said:
In your u substitution, have you applied the chain rule correctly?

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 :)
 

Similar threads

Replies
7
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K