Homework Help: 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?

21. Sep 21, 2010

judahs_lion

not really. im thinking insted of it substituting to udu, it would be something like u((1/2)du)?

22. Sep 21, 2010

Dick

Argh. Why don't you use the original substitution you suggested, u=ln(t^(1/2)). It works great. I promise you.

23. Sep 21, 2010

╔(σ_σ)╝

I promise too :).

24. Sep 21, 2010

judahs_lion

I just don't see it. Don't see du inside the original function.

25. Sep 21, 2010

Dick

I'm going to have to ask you again. If u=ln(t^(1/2)), what's du? Why don't you see du (or some constant multiple of it) in the original function???