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

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SUMMARY

The integral of ln²(6x) can be solved using integration by parts, where the first step involves setting u = ln²(6x) and dv = dx. The differentiation yields du = (2ln(6x))/x dx and v = x. The integration process leads to the expression xln²(6x) - 2∫ln(6x) dx, which requires further integration by parts. The final solution is xln²(6x) - 2xln(6x) + 2x + C, correcting the sign error in the constant term.

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  • Familiarity with logarithmic differentiation
  • Knowledge of the chain rule in calculus
  • Ability to manipulate integrals involving logarithmic functions
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banshee43
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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
 
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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 :)
 

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