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?

$$(2/3)\ln|3x|+c=(2/3)\ln|x|+(2/3)\ln(3)+c=2/3\ln|x|+c'$$ – Did Mar 15 '18 at 21:57 — Did, Mar 15 '18 at 21:57
Because the first $c$ is different than the second $c$. – Mehrdad Zandigohar Mar 15 '18 at 21:59 — Mehrdad Zandigohar, Mar 15 '18 at 21:59

score 2 · Accepted Answer · edited Mar 15 '18 at 22:22

2

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.

edited Mar 15 '18 at 22:22

Andrew Li

4,734

answered Mar 15 '18 at 22:05

Mohammad Riazi-Kermani

70,421

Thanks, should have spotted that really. The thought did not cross my mind! – Henry Lee Mar 15 '18 at 22:16
1

Don't worry. You are not the first one who is surprised by logarithms. – Mohammad Riazi-Kermani Mar 15 '18 at 22:20

score 0 · Answer 2 · answered Mar 15 '18 at 22:02

0

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

answered Mar 15 '18 at 22:02

J.G.

118,742

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

2 Answers2