<s>[INST] Integral of $\int \frac{\sqrt{9-x^2}}{x}$ I don't understand how the integral 
$$\int \frac{\sqrt{9-x^2}}{x}=\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$$
I keep getting $-3/x +c$ as the answer.
 [/INST] First thing - it is generally a bad idea to write $\int \dfrac{\sqrt{9-x^2}}{x}$ without the $dx$ part, it takes away the meaning of the integral.
So, solving $\int \dfrac{\sqrt{9-x^2}}{x} \cdot dx$
The key step is to consider the substitution $x = 3\sin \theta$. Thus, $dx = 3\cos \theta \cdot d\theta$
On substitution, the problem becomes $\int \dfrac{\cos \theta}{\sin \theta} \cdot 3\cos \theta \cdot d\theta$ = $\int 3 (\csc \theta - \sin \theta) \cdot d\theta$. 
From here, using $\int \csc \theta \cdot d\theta  = -\ln (\csc \theta + \cot \theta)$ and $\int \sin \theta \cdot d\theta = - \cos \theta$ and substituting back $x$ would give you the result
$\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$
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