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

In summary, the integral of ln^2(6x)dx can be solved using the integration by parts method, substituting u=ln^2(6x) and dv=dx. After applying the chain rule correctly, the integral can be simplified to xln^2(6x)-2xln(6x)+2x+C.
  • #1
banshee43
15
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
  • #2
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.
 
  • #3
is it because the -2 is distributed and not +2?
 
  • #4
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.
 
  • #5
yes! thank you so much... those simple mistakes will be the death of me!
 
  • #6
In your u substitution, have you applied the chain rule correctly?
 
  • #7
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 :)
 

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

1. What is integration by parts?

Integration by parts is a method of integration that allows us to find the integral of a product of two functions. It is based on the product rule of differentiation and involves breaking down a complicated integral into simpler integrals.

2. When should I use integration by parts?

Integration by parts is useful when the integrand (the function being integrated) contains a product of two functions, one of which can be easily integrated while the other can be easily differentiated. It is also helpful when attempting to integrate functions involving logarithms, trigonometric functions, and exponential functions.

3. How do I perform integration by parts?

The integration by parts formula is ∫u dv = uv - ∫v du, where u and v are the two functions being integrated and dv and du are their respective differentials. To use this formula, we choose u and dv in a way that makes the resulting integral ∫v du simpler to solve than the original integral ∫u dv. We then plug in the values for u, v, dv, and du and solve for the integral.

4. Are there any tricks or shortcuts for integration by parts?

There are a few tricks that can make integration by parts easier. These include choosing u and dv in a strategic way, using tabular integration for repeated integration by parts, and recognizing common patterns in the integrand.

5. What are the common mistakes to avoid in integration by parts?

One common mistake is choosing u and dv in a way that makes the resulting integral ∫v du more complicated than the original integral ∫u dv. Another mistake is forgetting to include the constant of integration when solving for the integral. It is also important to carefully differentiate and integrate each term in the formula, as mistakes in this step can lead to incorrect answers.

Similar threads

Solve the problem that involves implicit differentiation
    • Calculus and Beyond Homework Help
Replies
24
Views
1K
Find the derivative of given function and hence find its integral
    • Calculus and Beyond Homework Help
Replies
7
Views
1K
Integral using substitution x = -u
    • Calculus and Beyond Homework Help
Replies
12
Views
1K
Verify Green's Theorem in the given problem
    • Calculus and Beyond Homework Help
Replies
10
Views
680
Applying integration to math problems
    • Calculus and Beyond Homework Help
Replies
5
Views
877
How Do You Solve These Continuous Probability Problems?
    • Calculus and Beyond Homework Help
Replies
3
Views
943
How do you prove that ln(a^x) = xln(a) and a^x = e^xln(a) without using exponent rules?
    • Calculus and Beyond Homework Help
Replies
6
Views
793
How Do You Simplify and Analyze Taylor Polynomials for Higher Degree Functions?
    • Calculus and Beyond Homework Help
Replies
5
Views
1K
Integration problem using inscribed rectangles
    • Calculus and Beyond Homework Help
Replies
14
Views
645
Finding a definite integral from the Riemann sum
    • Calculus and Beyond Homework Help
Replies
2
Views
717

Hot Threads

Recent Insights

Back
Top