integral math problem?

Question

9
∫1/2x dx =
1

i know the answer is ln3, but i just can't seem to work it out that way, could someone explain it to me?

Answer 1

∫ 1/(2x) dx
(1/2) ∫ 1/x dx
= (1/2) [lnx]
now substitute the limits,
= (1/2) [ln(9) - ln(1)]
= (1/2) [ln(3^2) - 0]
= (1/2) [2ln3]
= ln(3)

Answer 2

I = ∫(1/2x) dx between 1 and 9
Let u = 2x
du = 2 dx
du/2 = dx
I = (1/2) ∫ (1/u) du between 2 and 18
I = (1/2) log u between 2 and 18
I = (1/2) (log18 - log 2)
I = (1/2) log 9
I = (1/2) log (3²)
I = (2) (1/2) log 3
I = log 3