In my maths class we are learning about indefinite integrals, this is the problem we were working on:

$$ \int \frac{1}{2x+1}dx $$

Using u-substitution we obtain:

$$ \frac{1}{2}\ln\left | 2x+1 \right | + C $$

But why does it not work to pull out a $\frac{1}{2}$ so that we don't have to do u-substitution

$$ \frac{1}{2}\int \frac{1}{x+\frac{1}{2}}dx $$

This yields a completely different result $$ \frac{1}{2}\ln \left |x+\frac{1}{2}\right | + C_1 \neq \frac{1}{2}\ln\left | 2x+1 \right | + C_2 $$

Pulling out the $\frac{1}{2}$ seems like a completely valid move, so why does it get a completely different result?

  • $\begingroup$ Is it the same constant of integration $C$ for each? $\endgroup$
    – Karl
    Feb 25, 2016 at 21:05
  • $\begingroup$ @Karl There's not initial condition so if they're not equivalent then no. $\endgroup$
    – BitNinja
    Feb 25, 2016 at 21:08
  • $\begingroup$ This was the root of the confusion. The answers were the same depending on the constant as @Spenser shows $\endgroup$
    – Karl
    Feb 25, 2016 at 21:10
  • $\begingroup$ It is correct, you can get "different" results because the integral is indefinite. Take limits in any "different result" and you will see that the integral is the same, no matter the form of the antiderivative. $\endgroup$
    – Masacroso
    Feb 25, 2016 at 21:26
  • $\begingroup$ @BitNinja Just because the constant of integration is 'different', which it will mostly be (if we were to have a definite integral), doesn't mean the ANSWERS are 'completely different'. They are just different forms. You should probably try with endpoints and then you'll see why they are in fact equivalent - as per Spenser's answer $\endgroup$ Feb 26, 2016 at 2:52

1 Answer 1


They are the same because $$\ln|2x+1|=\ln|2(x+1/2)|=\ln(2)+\ln|x+1/2|$$ and the constant $\ln(2)$ goes with the constant $C$.


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