4
$\begingroup$

This is a dumb question but I'd still like to confirm it here, I just haven't found any information about this in internet.

According to any table of integrals $\int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln(ax+b)$ . This doesn't work if if $b=0$, right? Because $\int\frac{1}{ax} \, dx$ should equal to $\frac{1}{a} \int \frac{1}{x} \, dx$ which would be $\frac{1}{a} \ln(x)$.

$\endgroup$
1
  • 1
    $\begingroup$ Try taking the derivative of the function you expect to be the answer. $\endgroup$ Mar 27, 2016 at 22:48

1 Answer 1

14
$\begingroup$

It still works, your result is the same as $\tfrac1a\ln(ax)$, up to a constant. Note that, $$\tfrac1a\ln(ax)=\tfrac1a(\ln a+\ln x)=\tfrac{\ln a}a+\tfrac1a\ln x.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .