Question

what is the integral of ln(sqt t)/t

Answer 1

Well,
\[ \ln \sqrt{t} = \ln t^{1/2} = \frac{1}{2} \ln t \]

Now see what to do?

Answer 2

I really don't. Don't is my u substitution ln\[\sqrt{t}\]

Answer 3

\[\ln \sqrt{t}\] is that the u sub
?

Answer 4

Let me ask you this. Can you evaluate this integral?
\[ \int \frac{\ln t}{t} dt \]

Answer 5

Because your integral is
\[ \int \frac{\ln \sqrt{t}}{t} dt = \int \frac{1}{2} \frac{\ln t}{t} dt = \frac{1}{2} \int \frac{\ln t}{t} dt \]

Answer 6

yeah, use by u sub... it will work

Answer 7

In this \( \it last \) integral, substitute u = ln t

Answer 8

well I don't understand how ln sqt (t) is the same as 1/2 ln(t)/t

Answer 9

\[ \ln \sqrt{t} = \ln t^{1/2} = \frac{1}{2} \ln t \]
Do you agree with that much?

Answer 10

i understand the t^1/2 but what property of logs makes t^1/2 the same as 1/2 ln t

Answer 11

Because
\[ \ln x^a = a \ln x \] for all values of \( a \).

You can see this easily enough for integer values of a: for example, if a = 3,
\[ \ln x^3 = \ln xxx = \ln x + \ln x + \ln x = 3 \ln x \]

Answer 12

ohhhhhhh, i do see. No one has ever pointed that out before. Thank you!!

Answer 13

Just to finish the argument for your particular case, we can also write
\[ \ln t = \ln \sqrt{t}\sqrt{t} = \ln \sqrt{t} + \ln \sqrt{t} \]
Hence
\[ 2 \ln \sqrt{t} = \ln t \]
and it must therefore be the case that
\[ \ln \sqrt{t} = \frac{1}{2} \ln t \]

Answer 14

Uh uh, another snag. When I u sub with u=ln t, I still end up with a u and a t, variable.

Answer 15

If u = ln t,
du = .... what?

Answer 16

or du/dt = ... what?

Answer 17

1/t dt, which is t du =dt

Answer 18

ohhhh, t= du/dt

Answer 19

du = dt/t, yes

Hence
\[ \int \frac{\ln t}{t} dt = \int \ln t \frac{dt}{t} = \int u \ du \]

Answer 20

wow. I wish I could see this more easily

Answer 21

Practice, practice, practice.

Answer 22

yes