Integrate 1/x = ln(x) vs ln(kx)

[Billy.Ljm]

If $k$ is a constant, I know
$\frac{d}{dx} \ln(x) = \frac{1}{x}$
$\frac{d}{dx} \ln(kx) = \frac{k}{kx} = \frac{1}{x}$

However, what about $\int\frac{1}{x}$.
I've been taught to use $\int\frac{1}{x} = \ln(x)$,
but wouldn't $\int\frac{1}{x} = \ln(kx)$ work as well.
And if this is true, there are an infinite number of integrals??

 

[Billy.Ljm]

Oh god, I just figured it out, right after posting this. I swear I was confounded for hours before this.
$ln(kx)$ works as well.
$ln(kx) = ln(x) + ln(k)$ and the $ln(k)$ would cancel out in any definite integrals.
But that would still mean there are an infinite number of indefinite integrals, which is a bit strange but perhaps possible.

 

[Mentallic]

Well, yes!

ln(kx) = ln(k)+ln(x)

and since k is just some constant, then ln(k) is also a constant. Remember that when you take the integral, you have to add +c at the end to denote that you can add any constant to the result and still have a correct solution.

So

$$\int\frac{1}{x}dx=\ln(kx)+c = \ln(x)+\ln(k)+c$$

for any constant c, but since c is any constant, then we can merge $\ln(k)+c$ into a new constant, say, c₂ which is the exact same result as if we left it as ln(x)+c to begin with.