Integration - problems with the range of x

Question

I am to calculate this integral (it doesn't come from any special equation or something - it's just an example):
$\int \frac{dx}{x(x^{2}-4)}$
I started with sums of unit fractions of course and I received:
$-\frac{1}{4} \int \frac{dx}{x} + \frac{1}{8} \int \frac{dx}{x+2} + \frac{1}{8} \int \frac{dx}{x-2}$
Now (let's skip the substitutions) I finally have:
$-\frac{1}{4} \log|x| + \frac{1}{8} \log|x+2| + \frac{1}{8} \log|x-2| + C = \frac{1}{8} \big(\log|x^{2}-4| - 2\log|x| \big) + C$
Wolfram says something else however:
$\frac{1}{8} \big(\log(4-x^{2}) - 2\log(x) \big) + C$
Where is the problem? I think it's something connected to the range of x (and abs of course) but I'm not sure if it's higly important.
I would appreciate any explanation :)

Answer 1

Put $t=x^2$ in $\int\frac{dx}{x(x^2-4)}$ gives

$$\frac{1}{8}\int (\frac{1}{t-4}-\frac{1}{t})dt$$

$$=\frac{1}{8}\ln( \frac{|x^2-4|}{x^2} )+C$$ $$=\frac{1}{8}\ln(|x^2-4|)-\frac{1}{4}\ln(|x|).$$

in each of the intervals: $(-\infty,-2),\;(-2,0),\;(0,2),(2,+\infty)$ where the function $x\mapsto \frac{1}{x(4-x^2)} $ is continuous.