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

[Dick]

yea, i don't know have to use symbols in here.

Ok, so you can integrate that, right?

 

[MysticDude]

I don't want to tell you the answer, but I'll give you a clue...that will pretty much help you get the answer. Now that we have everything with "u", we can integrate normally. As if you were integrating plain x. All you have to do is use the reverse power rule.

So what would you get if you use the reverse power rule on u?

 

[fizzynoob]

its a very simple integral after you use U substitution


\frac{1}{2}\int\frac{ln(t)}{t}dt

u = ln(t)
du = ?

 

[judahs_lion]

Ok, so you can integrate that, right?

2[{u^(3/2)}/(3/2)}]

so

[lnt^2] /3 ?

 

[MysticDude]

2[{u^(3/2)}/(3/2)}]

so

[lnt^2] /3 ?

from where did you get the 3/2. Remember that ln(√t) does not equal (ln(t))^(1/2). I think that the answer should have been u² => (ln(√t))².


I'm not 100% on this, but I think that this is the correct one.

 

[Dick]

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?

 

[judahs_lion]

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

 

[MysticDude]

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.