<s>[INST] Integral : $\int \frac{\sqrt{9-x^2}}{x^2}$ I have the following integral : 
$$\int \frac{\sqrt{9-x^2}}{x^2} = -\frac{\sqrt{9-x^2}}{x} - \arccos (\frac{x}{3})$$
But it seems like I must use $\arcsin$... Is there a difference between the two ?
 [/INST] First off the sign on the $\arccos(\frac{x}{3})$ is wrong.
I guess you could use the constant of integration to solve this. From the definitions of the inverse trig functions we know that 
$\arccos x = \frac{\pi}{2} - \arcsin x$
So your antiderivative can be written as
\begin{equation}
-\frac{\sqrt{9 - x^2}}{x^2} + \arccos(\frac{x}{3}) =  -\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})
\end{equation}
Since $-\frac{\sqrt{9 - x^2}}{x^2} + \frac{\pi}{2} - \arcsin(\frac{x}{3})$ and $-\frac{\sqrt{9 - x^2}}{x^2} - \arcsin(\frac{x}{3})$ have the same derivative(the constant $\frac{\pi}{2}$ does nothing) it means that they are both antiderivatives of your original function. This is why a $ + C$ is added at the end of the antiderivative's general form. 
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