Integrating with ln in denominator.

[Poop-Loops]

Ok, so it's been a while since I've had to integrate anything, much less something like this.

[tex]\int \frac{1}{n(1 + \ln{n})^{2/3}} dn[/tex]

I'm thinking u substition for ln(n) and then du becomes 1/n? But, since the ln(n) is in the denominator of a fraction raised to a power, wouldn't that mess with du? Or am I on the right track? I checked my calculus book and surfed the web, and couldn't find an integral for ln, so I can't see any other way...

If it helps, this is for a series check by integration to see if it's convergant or divergant, so it goes from nothing to infinity.

 

[koroljov]

Substitute u = ln(n), then substitute 1+u = v^3, and you simply have the integral of 3 dv

 

[moose]

You've got the right idea. Remember, a substitution is just that, substituting in another number or variable, so why would you be worried?

 

[Poop-Loops]

Because these things take a LONG time, and then I don't even know if I have the right answer unless I get stuck (then I know it's wrong).

Thanks. I'll play around with that tomorrow. I've been doing my homework for like 3 hours now, so I think I'll go to bed. :)

 

[moose]

Also, remember that you can always check your answer by taking the derivative of your answer and seeing if it turns out to be the integrand

EDIT: Just by looking at it, it seems that your answer will be very straighforward and take very little steps to arrive at.

EDIT2: koroljov: you shouldn't simply provide the answer in such a situation.