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This is a simple integral.

$$ \int \frac{1}{3x}dx $$

with an equally simple solution of

$$ \frac{1}{3}\ln|x| +c $$

My question is that if you chose to use u-substitution and used u = 3x, the solution appears to work out as follow:

$$ \int \frac{1}{u} \frac {du}{3} $$ $$ \frac{1}{3} \ln|3x|+c $$

which seems correct as well. Is this in fact correct? The 2 graphs appear nothing alike.

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    $\begingroup$ both answers are correct and are the same, just write $C:=\ln 3+K$ $\endgroup$
    – Masacroso
    Commented Apr 23, 2018 at 19:11
  • $\begingroup$ By the way, the graphs are actually pretty similar. $\endgroup$
    – Javier
    Commented Apr 24, 2018 at 14:43

3 Answers 3

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It's the same since you have a constant of integration.. $$ \frac{1}{3} \ln|3x|+c= \frac 13\ln |x|+ \underbrace{\frac{1}{3} \ln|3|+c}_{ \text { is a constant } }=\frac{1}{3} \ln|x|+K$$

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    $\begingroup$ Thanks guys. Completely forgot about expanding the log. $\endgroup$
    – Deeves
    Commented Apr 23, 2018 at 19:18
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    $\begingroup$ yw @Deeves its the constant of integration that absorb the $ \log(3)$ $\endgroup$
    – user577215664
    Commented Apr 23, 2018 at 19:20
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    $\begingroup$ Good points are living here! +1 $\endgroup$
    – Mikasa
    Commented Aug 13, 2018 at 15:07
  • $\begingroup$ thank you @ResidentDementor $\endgroup$
    – user577215664
    Commented Aug 13, 2018 at 15:10
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Hint: use that $$\ln(3x)=\ln(3)+\ln(x)$$

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If you choose the substitution $u=3x\implies du=3\,dx$, then $$\int\frac{dx}{3x}=\int\frac{3\,dx}{3\cdot3x}=\frac13\int\frac{du}{3u}=\frac13\ln|3u|+C=\frac13\ln3+\frac13\ln|u|+C=\color{red}{\frac13\ln|u|+C_1}$$ where $C_1=\frac13\ln3+C$ is another constant.

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