## jotopia34 Group Title What is the integral of ln(sqrt t)/t, stuck again! one year ago one year ago

1. tkhunny Group Title

What have you tried?

2. jotopia34 Group Title

I will type in what I've done right now

3. tkhunny Group Title

Sweet!

4. amoodarya Group Title

5. tkhunny Group Title

?? Is that the same problem?

6. jotopia34 Group Title

$u=\sqrt{t}$ 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 Group Title

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

8. tkhunny Group Title

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

9. slaaibak Group Title

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

10. jotopia34 Group Title

what is that property again please? in general format?

11. slaaibak Group Title

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

12. jotopia34 Group Title

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

13. slaaibak Group Title

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

14. jotopia34 Group Title

yes, I believe so

15. slaaibak Group Title

shot

16. jotopia34 Group Title

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 Group Title

18. jotopia34 Group Title

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

19. slaaibak Group Title

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 Group Title

yes!! You are all geniuses!

21. slaaibak Group Title

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

22. jotopia34 Group Title

insomia is the curse of genius