# Indefinite integratrion

1. Sep 21, 2010

### judahs_lion

1. The problem statement, all variables and given/known data
integrate lnsqrt(t)/t , dt

2. Relevant equations

3. The attempt at a solution

This is the most i can come up with.

u = lnsqrt(t), so du = 1/(2sqrt(t))dt , so dt = 2du/sqrt(t)

Im stuck @ this point

2. Sep 21, 2010

### ╔(σ_σ)╝

Hint: The logarithm function has an interesting property.

$$ln\left(a^{b} \right) = bln\left(a \right)$$

3. Sep 21, 2010

### Dick

du isn't 1/(2*sqrt(t))*dt. You are forgetting to use the chain rule. You can also get the correct differential by applying the rules of logs to ln(sqrt(t))=ln(t^(1/2)).

4. Sep 21, 2010

### judahs_lion

so i would integrate (lnt^(1/2))/t ?

5. Sep 21, 2010

### Dick

Sure. Isn't ln(sqrt(t))/t=ln(t^(1/2))/t?

6. Sep 21, 2010

### judahs_lion

Ok, now i'm not sure what to take as u.

7. Sep 21, 2010

### ╔(σ_σ)╝

Yes.
But follow Dicks' suggestion also. It would be helpful to you. Your original method was correct ,however, as already pointed out you weren't differentiating correctly.

8. Sep 21, 2010

### Dick

??? Take u the same as you originally chose. u=ln(sqrt(t))=ln(t^(1/2)). They are the same thing. You just have to compute du correctly. That's where your first attempt went wrong. The du was wrong.

9. Sep 21, 2010

### judahs_lion

Yea i think i integrated rather then differentiated to get du. so u = 1/2lnt , du = 1/2(1/t)dt?

10. Sep 21, 2010

### judahs_lion

actually du = 1/t

11. Sep 21, 2010

### judahs_lion

that would make give me 1/2lnt (1/t), which is udu +C. so it breaks down to u^2/2 + C?

12. Sep 21, 2010

### MysticDude

Last edited by a moderator: Apr 25, 2017
13. Sep 21, 2010

### Dick

Sure. You could have also done it with ln(sqrt(t)). (ln(sqrt(t))'=(1/sqrt(t))*(dsqrt(t)/dt)=(1/sqrt(t))*(1/(2*sqrt(t)))=1/(2t). That's the chain rule. BTW if you aren't going to use TeX you should probably use a few more parentheses. 1/2lnt could mean (1/2)ln(t) or 1/(2ln(t)) etc.

14. Sep 21, 2010

### MysticDude

Okay I'll think about that next time.

15. Sep 21, 2010

### judahs_lion

Last edited by a moderator: Apr 25, 2017
16. Sep 21, 2010

### MysticDude

No all it is is just normal differentiation. See all you have to do if factor out the constant( 1/2 in this case). So it would look like $$(\frac{1}{2}) * (\frac{d}{dt})(ln(t)) = \frac{1}{2t}$$

17. Sep 21, 2010

### judahs_lion

but i thought the 1/2 would be eliminated because it is a constant

18. Sep 21, 2010

### MysticDude

If you can factor out a constant, then do so. Plus if you look at the link that I gave you. There should be a "Show Steps" link. It says to factor out the constant.

AND even if you do it by using the product rule (I'm doing this so you can see that the 1/2 stays)

$$(\frac{1}{2}[(\frac{d}{dt})(ln(t))] + (\frac{d}{dt})(\frac{1}{2})(ln(t))$$
We have $$\frac{1}{2}*\frac{1}{t} + 0(this-is-the-derivative-of \frac{1}{2}) * ln(t)$$
so this shows that the derivative of $$\frac{1}{2} * ln(t) = \frac{1}{2t}$$

I hope you understand!

19. Sep 21, 2010

### judahs_lion

Thank you

20. Sep 21, 2010

### Dick

Ok, so after all that fuss, you can do the integral, right?