What's wrong with my approach - reciprocal integral I want to integrate $\int \frac{1}{2x}dx$
The way I would solve it: $$\frac{1}{a}f(ax + b)$$ where $$f(x) = \ln x$$
My answer $\frac{1}{2}\ln(2x) + k$
The correct way seems to be to factor out $\frac{1}{2}$ from the initial integral, leaving $\int \frac{1}{x}$. The correct answer $\frac{1}{2}\ln(x) + k$
Where did I go wrong?

You are not wrong; however, your answer is just not simplified.
To check that your answer is correct, differentiate both your answer and the "correct" answer:
$$\frac{d}{dx} (\frac{1}{2} ln(2x) + k) = \frac{1}{2x}$$
$$\frac{d}{dx} (\frac{1}{2} ln(x) + k) = \frac{1}{2x}$$
To simplify:
$\frac{1}{2} ln(2x) + k = \frac{1}{2} (ln(x) + ln(2)) + k$
$=> \frac{1}{2} ln(x) + \frac{1}{2} ln(2) + k = \frac{1}{2} ln(x) + (\frac{1}{2} ln(2) + k)$

$(\frac{1}{2} ln(2) + k)$ is another constant, so let's just call that C.

$=> \frac{1}{2} ln(2x) + k = \frac{1}{2} ln(x) + C$
C and k are just constants and do not affect the rate of change of the function; therefore, it doesn't matter what constant you add at the end of your equation as long as it is indeed constant.
