Why is the integral of $0.5/x = 1/2 \ln x$ and not $1/2 \ln 2x$? When you calculate $\int\frac{1}{2x}dx$ you get $\frac{1}{2}\ln(x)$ and when you calculate $\int\frac{1}{2x}dx$ you don't get $\frac{1}{2}\ln(2x)$.
$\frac{1}{2x}$ is the same as $\frac{1}{2x}$
why do you get different answers?                                 
 A: I think that @5xum has already said it, but to make it super clear:
The integral of $1 / (2x)$ is not equal to $\frac{1}{2}\ln(x)$ or $\frac{1}{2}\ln(2x)$. you have
$$
\int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C.
$$
As noted in the other answer and in the comments, this $+C$ isn't just a pretty thing. It is very important because without it, the answer is wrong. The example that you have given shows exactly why it is important. So you have, for example,
$$
\int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C_1 = \frac{1}{2}\ln(2x) + C_2
$$
where $C_1$ and $C_2$ are different constants. The reason that this is true is because of the definition. Remember that we say that 
$$
\int f(x)\; dx = F(x) + C
$$
exactly when $F'(x) = f(x)$. And indeed you have that 
$$\begin{align*}
\frac{d}{dx} \frac{1}{2}\ln(x) = \frac{d}{dx} \frac{1}{2}\ln(2x) = \frac{1}{2x}.
\end{align*}
$$
A: $$\frac{1}{2}\ln(2x) = \frac{1}{2}(\ln 2 + \ln x) = \frac{\ln2}{2} + \frac12\ln x$$
When you integrate $$\frac{\frac12}{x},$$ you get $\frac12 \ln x + C$ ($C$ is the constant you are always nagged about!), not just $\frac12\ln x$
