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.
 A: Differentiate:
$$\left(\sqrt{9-x^2}-3\log(\sqrt{9-x^2}+3)+3\log x+C\right)'=$$
$$=-\frac x{\sqrt{9-x^2}}+\frac{3x}{\sqrt{9-x^2}}\cdot\frac1{\sqrt{9-x^2}+3}+\frac3x=$$
$$=\frac x{\sqrt{9-x^2}}\left(-1+\frac3{\sqrt{9-x^2}+3}\right)+\frac3x=$$
$$=\require{cancel}\frac x{\cancel{\sqrt{9-x^2}}}\frac{-\cancel{\sqrt{9-x^2}}}{\sqrt{9-x^2}+3}+\frac3x=\frac{9-x^2+3\sqrt{9-x^2}}{x(\sqrt{9-x^2}+3)}=$$
$$\frac{\sqrt{9-x^2}\left(\sqrt{9-x^2}+3\right)}{x(\sqrt{9-x^2}+3)}=\frac{\sqrt{9-x^2}}x$$
so they are right, you are wrong.
A: Hint:  if you put $x=3sin(t)$  with $t\in (0,\pi)$  then
your function becomes
$\frac{3\sqrt{1-sin^2(t)}}{3sin(t)}$
and $dx$ becomes $3cos(t)dt$.
your integral will be 
$\int \frac{3cos^2(t)sin(t)}{1-cos^2(t)}dt$
which can be finished by putting 
$u=cos(t)$
and converting the fraction obtained into its partials.
A: 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
$\sqrt{9-x^2}-3 \ln(\sqrt{9-x^2}+3)+3 \log(x)+c$
