Integrating ln(sqrt(t)/t) using u-substitution

  • Thread starter judahs_lion
  • Start date
  • Tags
    Indefinite
In summary: It's du=(1/(2t))*dt, ok? Can you show what you get when you use that substitution into the original... or if you do the substitution, you'll get (1/u)*du = (1/(2t))*dt, right?In summary, the integral of ln(sqrt(t))/t, dt can be solved by substituting u=ln(t^(1/2)) and using the property of logarithms to get du=1/(2t). This results in the integral becoming 1/u*du=(1/2t)*dt, which can then be solved using
  • #36
judahs_lion said:
2[{u^(3/2)}/(3/2)}]

so

[lnt^2] /3 ?

You aren't making any sense. Why don't you try this again in the morning?
 
Physics news on Phys.org
  • #37
Dick said:
You aren't making any sense. Why don't you try this again in the morning?

Was just bout to say that. Got homework due for two other class tommorow. Thanx for all your help @ everyone
 
  • #38
Remember to use Wolfram Alpha. It is a VERY useful to for math! I use whenever I'm stuck. I mean it even has a "show steps" button. It is the Google of Math. Well, I'm done too. G'night people.
 
Back
Top