Integration of fractions and cancellation

Question

How come $$\int\frac{3}{6+3x}\mathrm{d}x=\mathrm{ln}(2+x)+C$$ and not $\ \mathrm{ln}(6+3x)+C$ ?

I understand you can cancel the fraction but why should this make a difference?

I've tried searching through all my notes on this and online but the terms I am using do not return anything useful.

the difference between the two antiderivatives is constant; $\ln(6+3x) - \ln(2+x) = \ln 3 .$ — abel, Jun 08 '15 at 21:49

score 5 · Accepted Answer · answered Jun 08 '15 at 21:19

5

$$\ln(6+3x)+C=\ln(3[2+x])+C=\ln(2+x)+(\ln 3+C)$$

In other words, both answers are correct, they just use different arbitrary constants.

answered Jun 08 '15 at 21:19

Rory Daulton

score -2 · Answer 2 · edited Jun 18 '15 at 03:24

-2

It is divided by $3$ and thus follows $\log(2+x)$

edited Jun 18 '15 at 03:24

Joaquin Liniado

answered Jun 08 '15 at 21:46

user246780

and Log(2+x)*3=Log(2+x)+Log3 – user246780 Jun 08 '15 at 21:48
1

This answer is not particularly clear... Also you can edit your answer, if you want to add things to it. – Jun 08 '15 at 22:16

2 Answers2