<s>[INST] Why do I get two different indefinate integrals for the same integrand? (u-substitution) Using u-substitution I'm able to integrate the following two integrands: 
$$\int \frac{1}{2x+3}dx = \frac{1}{2}\ln|2x+3|+C$$
$$\int \frac{1}{2x+5}dx = \frac{1}{2}\ln|2x+5|+C$$
Following this pattern I would assume:
$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|2x+4|+C$$
However when I enter this expression into wolfram alpha I get:
$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|x+2|+C$$
I can even arrive at this result myself by factoring out the $\frac{1}{2}$ before integration like this:
$$\int \frac{1}{2}\cdot\frac{1}{x+2}dx = \frac{1}{2}\int \frac{1}{x+2}dx$$
But I don't understand how this can be the case? How can two different techniques give me different results? Is there a property of logs that I don't understand? 
I would think that
$$\frac{1}{2}\ln|x+2|+C \neq \frac{1}{2}\ln|2x+4|+C$$
Or are they actually equal somehow? PLEASE help me understand what's going on here.
 [/INST] Because $$\frac{1}{2}\ln|2x+4|+C=\frac{1}{2}\ln|x+2|+\frac{1}{2}\ln2+C,$$
where $\frac{1}{2}\ln2+C$ is a constant. 
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