Integrating, probably by parts

  • Thread starter Thread starter mbrmbrg
  • Start date Start date
  • Tags Tags
    Integrating parts
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
mbrmbrg
Messages
486
Reaction score
2
I have the expression [tex]\int{x(\ln{x})^3dx}[/tex]
I thought I had a quick way to integrate by parts but it turned out that I had accidentally evaluated [tex]\int{x\ln{x}dx}[/tex] instead.
Revisiting [tex]\int{x(\ln{x})^3dx}[/tex], I wanted to start by making a strange substitution, wherein u=ln(x), du=1/x dx, and x=e^u. This meant that when I rewrote the integral, instead of multiplying dx by a constant to get it to be du, I multiplied it by x (which in this case was e^u). Is that allowed? Because I got a very different, much uglier answer than the book's.

I'd appreciate any comments, whether on my weird "method" or on a more standard approach to evaluating [tex]\int{x(\ln{x})^3dx}[/tex]
 
Physics news on Phys.org
Try integration by parts with u = (ln(x))^3 and dv = x dx
 
Last edited:
Your substitution method should work fine. Your should be integrating [tex]\int{Exp[2u] u^3du}[/tex]. If you do it by integration by parts, you will need to do it 3 times.
 
wurth_skidder_23 said:
Try integration by parts with u = (ln(x))^3 and dv = x dx
thanks, that got me the book's answer!
 
And yes, the other way does work also. Nifty!