Intergration by Parts (IbP) problem

  • Context: Undergrad 
  • Thread starter Thread starter Ravenatic20
  • Start date Start date
  • Tags Tags
    Intergration parts
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 4K views
Ravenatic20
Messages
30
Reaction score
0
[tex]\int_{1} ^{2e} x^2(ln x)^{2} dx[/tex]

I need to solve this using IbP. I made the following:

[tex]u = (ln x)^2[/tex]

[tex]du = (\frac{2 ln x}{x}) dx[/tex]



[tex]dv = x^2 dx[/tex]

[tex]v = \frac{x^3}{3}[/tex]

So I get:
[tex](ln x)^2 (\frac{x^3}{3}) \|_{1} ^{2e}[/tex][tex]- \int_{1} ^{2e} (\frac{x^3}{3}) 2 ln x dx[/tex]
(not sure how to make this look right)

Is this right?
Where do I go from here? Thanks
 
Physics news on Phys.org
Integrate by parts again.
 
So what I have is right so far? I just IbP on the RHS?
 
Yup, if you've integrated ln(x) before, it should be obvious that you can integrate it again.