TITLE: Can I do this when integrating 1/x to get rid of the absolute value?
QUESTION [0 upvotes]: I'm studying PDEs from a textbook and what comes up all the time is an ODE containing the integration of $1/x$. Now, if we are not sure that x is positive we can't get away from the absolute value in the integration. So, up to this point I would do this : 
$$\frac{dx}{x}=dy$$
$$\ln|x|=y+C_1$$
where $C_1$ is a constant. The author,however, in such cases, does the following: 
$$\frac{dx}{x}=dy$$
$$=>\ln (C_1x)=y$$
I know how you can get the constant inside the log but how does the absolute value vanish? 
In this problem nothing is stated about the value of x and the log includes no absolute value.

REPLY [2 votes]: Since the $C_1$ can be positive or negative, it can absorb the sign on $x$.  One just has to make sure that $C_1$ and $x$ have the same sign.
I would have taken a different approach.  From
$$\ln |x|= y+C$$
exponentiate both sides to get
$$| x | = e^{y+C} = =e^ye^C = Ae^y$$
where $A=e^C$ is a positive constant.  Then
$$x = \pm Ae^y$$
so all we need to do is allow $A$ to be positive or negative (or zero)
and it "absorbs" the plus-or-minus sign.  Now you have
$$x= Ae^y$$
with no absolute values.  Now you can put it in the author's form, if you like.
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