## jotopia34 2 years ago What is the integral of ln(sqrt t)/t, stuck again!

1. tkhunny

What have you tried?

2. jotopia34

I will type in what I've done right now

3. tkhunny

Sweet!

4. amoodarya

5. tkhunny

?? Is that the same problem?

6. jotopia34

$u=\sqrt{t}$ du= 1/2 t^-1/2 dt du=1/2*1/sqr(t) dt But I know its wrong because I still can't cancel anything

7. amoodarya

i think it was integral ln sqrt x dx isnt it ? or it is int ln(sqrtx)/x dx ?

8. tkhunny

Why would you use a square root substitution when you can use a logarithm property?

9. slaaibak

yeah.. ln(sqrt(t)) = 1/2 ln t

10. jotopia34

what is that property again please? in general format?

11. slaaibak

$\ln(x^a) = a \times \ln(x)$

12. jotopia34

ohhhh, now I see, the square root is 1/2 as exponent. Thus, that is my (a). Thank u!!

13. slaaibak

Cool :) do you know how to go from here?

14. jotopia34

yes, I believe so

15. slaaibak

shot

16. jotopia34

is the answer 1/4 * (lnx)^2+C? I think its wrong. I saw a post from another guy and I don't get what he wrote. Slaaibak, you make excellent sense. Can you post the solution steps? Id be eternally grateful...

17. amoodarya

18. jotopia34

Amoodarya, your last posting is the final answer I got, so I think I got it right....

19. slaaibak

Sure. $\int\limits {1 \over 2} {\ln t \over t} dt$ Let's set k = ln t dk = 1/t dt so the integral becomes; ${1 \over 2} \int\limits k dk = {1 \over 4 }k^2 + C = {1 \over 4}(\ln t)^2 + C$

20. jotopia34

yes!! You are all geniuses!

21. slaaibak

haha, if only :( hope it helped. thank you for the fun, makes insomnia a bit better.

22. jotopia34

insomia is the curse of genius