Where is the mistake in computation of $\int\!\frac{\mathrm dx}{10x-3}\;?$

Question

$\displaystyle\int\frac{\mathrm dx}{10x-3}=$ ?

Obviously, if we say $\,u=10x-3\,,\,$ then integration becomes

$\displaystyle\frac{1}{10}\int\frac{\mathrm du}{u}=\frac {1}{10}\ln(10x-3)+c$

However, I approach this question like

$\displaystyle\frac{1}{10}\int\frac{\mathrm dx}{x-\frac{3}{10}}\;,\;\;$ then $\;\;\dfrac{1}{10}\ln\left(x-\dfrac{3}{10}\right)+C.$

What am I missing ?

It is the same up to an additive constant $\frac{\ln 10}{10}.$ — Anne Bauval, May 25 '23 at 16:02
I am sorry but how can you see the difference between them is $\ln10 /10$ — Fuat Ray, May 25 '23 at 16:08
The two results are equal, in fact $,C=c+\frac1{10}\ln10,.;$ — Angelo, May 25 '23 at 16:13
$$\frac {1}{10}\ln(10x-3)+c=\frac {1}{10}\ln(10(x-3/10))+c=\frac {1}{10}(\ln(10)+\ln (x-3/10))+c.$$ — Anne Bauval, May 25 '23 at 16:15
Thank you very much for clarification. Now I fully understand — Fuat Ray, May 25 '23 at 16:17
Btw, beware that an antiderivative of $\frac1u$ is not $\ln u$ but $\ln|u|.$ — Anne Bauval, May 25 '23 at 16:21
Also don't forget that the standard antiderivative of 1/x is ln|x| + C (not ln(x) + C). — Dan Asimov, May 25 '23 at 16:22

score 0 · Answer 1 · answered May 25 '23 at 16:28

$$\ln(\frac{a}{b})=\ln(a)-\ln(b)$$ so $\frac{1}{10}\ln(x-\frac{3}{10})+C=\frac{1}{10}\ln(\frac{10x-3}{10})+C=\frac{1}{10}\ln(10x-3)-\frac{1}{10}\ln(10)+C$ but $-\frac{1}{10}\ln(10)+C$ is constant so let it equal $K$ then $$\frac{1}{10}\ln(10x-3)-\frac{1}{10}\ln(10)+C=\frac{1}{10}\ln(10x-3)+K$$

Where is the mistake in computation of $\int\!\frac{\mathrm dx}{10x-3}\;?$

1 Answers1