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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 $\frac{1}{a} \ln(x)$.

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    $\begingroup$ Try taking the derivative of the function you expect to be the answer. $\endgroup$ – YoTengoUnLCD Mar 27 '16 at 22:48
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It still works, your result is the same as $\tfrac1a\ln(ax)$, up to a constant. Note that, $$\tfrac1a\ln(ax)=\tfrac1a(\ln a+\ln x)=\tfrac{\ln a}a+\tfrac1a\ln x.$$

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