What is the integral of ln(sqrt t)/t, stuck again!

Question

tkhunny · Answer

What have you tried?

jotopia34 · Answer

I will type in what I've done right now

tkhunny · Answer

Sweet!

amoodarya · Answer

attachment

tkhunny · Answer

??  Is that the same problem?

jotopia34 · Answer

$$u=\sqrt{t}$$
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

amoodarya · Answer

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

tkhunny · Answer

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

slaaibak · Answer

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

jotopia34 · Answer

what is that property again please? in general format?

slaaibak · Answer

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

jotopia34 · Answer

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

slaaibak · Answer

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

jotopia34 · Answer

yes, I believe so

slaaibak · Answer

shot

jotopia34 · Answer

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...

amoodarya · Answer

attachment

jotopia34 · Answer

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

slaaibak · Answer

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;

$${1 \over 2} \int\limits k dk = {1 \over 4 }k^2 + C = {1 \over 4}(\ln t)^2 + C$$

jotopia34 · Answer

yes!! You are all geniuses!

slaaibak · Answer

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

jotopia34 · Answer

insomia is the curse of genius