ok
what is the integral of ( 1/2t)
i did a u sub
so u = 2t
du = 2
2 * ln2t
is this correct?
You don't need any kind of u or whatever conversion. It is a simple basic integral. All you need to know is:
int(1/t) = ln(t). THAT'S IT.
1/2 is just a constant. So the answer of your questions is:
int(1/2t)=(1/2) int(1/t)=(1/2) ln(t). AND THAT'S IT.
Which may be different than ln(t/2).
a*ln(t)=ln(a*t) is NOT always true.
So (1/2)*ln(t) = ln((1/2)*t) is NOT always true.
-O
If for some reason you decided to use a u-substitition:
$\displaystyle {\color{red}u = 2t} \ \Rightarrow \ {\color{blue}\tfrac{1}{2}du = dt}$
We then get: $\displaystyle \int \frac{1}{{\color{red}2t}} \ {\color{blue}dt} \ = \ \frac{1}{2} \int \frac{{\color{blue}du}}{{\color{red}u}} \ = \ \frac{1}{2} \ln u + C \ = \ \frac{1}{2} \ln 2t + C$
Looks a bit different. But recall: $\displaystyle \ln (ab) = \ln a + \ln b$
So: $\displaystyle \frac{1}{2} \ln 2t + C = \frac{1}{2} \ln 2 + \frac{1}{2} \ln t + C$
But $\displaystyle \tfrac{1}{2}\ln 2$ is nothing more than a constant. So we'll combine it with $\displaystyle C$ to make a new constant of integration which we'll call $\displaystyle K$.
i.e. $\displaystyle \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \ln 2t + C \ = \ \frac{1}{2} \ln t + K$
In other words, $\displaystyle \tfrac{1}{2} \ln 2t$ and $\displaystyle \tfrac{1}{2} \ln t$ are two legit antiderivatives of $\displaystyle \tfrac{1}{2t}$