Integral of 1/(8+2x^2) I have been following a rule saying that 
$$\int{\frac{1}{a^2+u^2}}dx = \frac{1}{a}\tan^{-1}(\frac{u}{a})+c$$
The question is asking for the interval of 
$$\frac{1}{8+2x^2}$$
Following that rule
$$a=\sqrt8$$
$$u=2x$$
So
$$\int{\frac{1}{8+2x^2}}dx = \frac{1}{\sqrt8}\tan^{-1}(\frac{2x}{\sqrt{8}})+ c$$
This seemed to have worked in the past but Wolfram is saying it is equal to 
$$\frac{1}{4}\tan^{-1}(\frac{x}{2})+ c$$
They have used the rule stating $$\int{\frac{1}{u^2+1}}dx=\tan^{-1}(u)+c$$
And they factor out the constants of the equation to get to that form.
My question is, is the way I am doing it okay, or should I be adopting the other method?

Both rules yield the same result if do things properly.
In this case you forgot to change $du = 2dx$;
Write explicitly the integral after the change of variable, without forgetting to change  the differential 
