Is $\int \frac2{3x}dx$ equal to $\frac23\ln|3x|+c$ or to $\frac23\ln|x|+c$?

Question

The question which caused an issue is $$\int \frac 2{3x}\, dx$$ The way I would always approach this would be to bring the $2/3$ outside as a constant and do $2/3\cdot\int (1/x)\, dx$, giving $2/3\cdot\ln|x|+c$.

Another method that a lot of people remember is that the integral of $f'(x)/f(x)=\ln(f(x))$.

In this case you would correct the top to equal three and place $2/3$ before the integral. This gives $2/3\int \frac{3}{3x} \, dx$, even though we can simplify $\frac 3{3x}$ to $1/x$. this then gives $2/3\cdot \ln|3x|+c$. So why do these two results come out differently, am I missing something here?

Answer 1

You are correct in noticing that $$\frac {2}{3x} = \left(\frac {2}{3}\right) \frac{1}{x}= \left(\frac {2}{3}\right) \frac{3}{3x}$$

Your confusion is due to the constant which appears after integration.

Note that $$\ln(3x)=\ln 3 + \ln(x) = \ln(x)+C $$

Thus both answers are as good as gold.

Answer 2

Both are correct for a suitable choice of $c$. By the way, $c$ is locally constant, but its value can differ either side of $x=0$.