On differentiating $F(x)=\ln(2x)$

Question

If we differentiate $F(x)=\ln(2x)$ we will get $F'(x) =\dfrac2{2x}$ after the shortcut $F'(x)=\dfrac1x$, right?

Now if we integrate $F'(x)$ we will get $F(x)=\ln(x)$ but also $F(x)=\ln(2x)$. This means $\ln(2x)=\ln(x)$. What happened?

Answer 1

When you integrate, there is an arbitrary constant $C$ added. So you only get $$\ln(2x)=\ln(x)+C.$$

Answer 2

But$$F(x)=\ln(2x) =\ln 2 + \ln x$$ So when you integrate back you get $$\ln x + C $$ where the constant $C$ should be the $\ln 2$ term.

Answer 3

The antiderivatives of $1/x$ are not uniquely determined ! They are given by

$$\ln x +C.$$

Forthermore $ \ln (2x)= \ln 2 + \ln x.$

Can you proceed ?