Integral of $\frac{1}{2x}$

Question

The integral of $\frac{1}{2x}$ is $\frac{\ln(x)}{2}$, but can't it also be $\frac{\ln(2x)}{2}$ or $\frac{\ln(3x)}{2}$?

Is there a special reason for $\ln(Ax)$ to have identical derivatives?

Answer 1

Differentiate the following identity $$\ln(Ax)=\ln(A)+\ln(x).$$

Answer 2

You might also consider it a consequence of the chain rule. For nonzero constants $A$,

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \ln(Ax) &=\frac{1}{Ax}\frac{\mathrm{d}}{\mathrm{d}x}(Ax)\\ &=\frac{1}{Ax}A\\ &=\frac{1}{x}. \end{align*}

Answer 3

$$\int \frac{dx}{2x} = \frac{1}{2} \int \frac{2dx}{2x}$$ put $ 2x=u \implies 2dx=du$ $$= \frac{1}{2} \int \frac{du}{u}$$ $$= \frac{1}{2} ln(u)$$

$$= \frac{ln(2x)}{2}$$