**58** 0

RobusEtCeleritas ln(2x) = ln(2) + ln(x), from logarithm identities.
Now you can differentiate term-by-term, and because ln(2) is a constant, you get
d/dx ln(2x) = d/dx ln(x).
When you go the other way and do an indefinite integral, there is a constant of integration. So the indefinite integral of dx/x is ln(x) + C.
This corresponds to an infinite family of functions, all of which have derivative 1/x.
ln(2x) is one member of this family.

**6** 0

rubseb Suppose f'(x) is the derivative of the function f(x) (i.e. f'(x) = d/dx f(x)). Now let's ask: is f'(x) *only* (uniquely) the derivative of f(x), or of other functions as well? The answer is that, since the derivative of a constant is 0, any function g(x) = f(x) + C also has the derivative f'(x). We can add (or subtract) any constant from f(x) and its derivative won't change. This is why, when we compute the indefinite integral of a function, we say that it is some function plus an unknown constant C.
In your example, the derivative of f(x) = ln(2x) is f'(x) = 1/x. If we compute the indefinite integral of f'(x), we get F(x) = ln(x) + C. Note, though, that the original function f(x) can be rewritten as f(x) = ln(2) + ln(x) (because log(ab) = log(a) + log(b)). So if we set C equal to ln(2) in F(x), we get back the original function f(x).
So the 2 isn't really lost, but because ln(2) is a constant, it is absorbed into the unknown constant 'C' in the indefinite integral.

**4** 0

efrique Into the constant.
ln(2x) = ln(x) + ln(2)
ln(3x) = ln(x) + ln(3)
etc.
when you integrate (find the antiderivative) you don't get ln(x), you get ln(x) + C
if you only state what the derivative is without giving any value for the original function, you can't tell what the constant should be; ln(kx) for every k will have the same derivative.