Integrating 1/3x: Is it 1/3 ln 3x or 1/3 ln x?

[Mattofix]

the intergral of 1/3x either equals 1/3 ln 3x or 1/3 ln x - i don't know which one is correct though because can't you take the 1/3 out of the intergral and then you get 1/3 ln x - so confused - i should really know this...

 

[arildno]

It doesn't matter which one you use, since they only differ by a constant:
$$\frac{1}{3}\ln(3x)=\frac{1}{3}\ln(x)+\frac{1}{3}\ln(3)$$
where the last term is simply a constant, as claimed.

 

[HallsofIvy]

the intergral of 1/3x either equals 1/3 ln 3x or 1/3 ln x - i don't know which one is correct though because can't you take the 1/3 out of the intergral and then you get 1/3 ln x - so confused - i should really know this...

You are wrong about "can't take the 1/3 out of the intergral" ("intergral": Boston accent?). $\int 1/(3x)dx= \int (1/3)(1/x)dx= (1/3)\int (1/x)dx= (1/3) ln x+ c$

It doesn't matter which one you use, since they only differ by a constant:
$$\frac{1}{3}\ln(3x)=\frac{1}{3}\ln(x)+\frac{1}{3}\ln(3)$$
where the last term is simply a constant, as claimed.

Of course, you could also let u= 3x so that du= 3 dx and the integral becomes
$$\int (1/(3x)) dx= \int (1/u)(du/3)= (1/3)ln(u)+ C= (1/3)ln(3u)+ C$$
As arildno points out, since ln(3u)= ln(u)+ ln(3), this just becomes
(1/3)ln(u)+ (1/3)ln(3)+ C and (1/3)ln(3)+ C is just a different "constant of integration".