Solution of $\int \frac{1}{x^2 \sqrt{x^2+9}}dx$ I'm new of the site. I must solve this exercise:
$$\int \frac{1}{x^2 \sqrt{x^2+9}}\,dx$$
I tried every substitution, but I didn't reach that I want.
Can you help me, please?

Let $x = 3\tan(u).$ Therefore $\frac{dx}{du} = 3\sec^2{u} \implies dx=3\sec^2{u}\ du$. So, our integral becomes 
$$\int\frac{3\sec^2{u}}{9\tan^2(u)\sqrt{9\tan^2(u)+9}}du= \int\frac{\sec^2{u}}{9\tan^2(u)\sqrt{\tan^2(u)+1}}du$$
$$=\int\frac{\sec{u}}{9\tan^2(u)}du$$
$$=\frac{1}{9}\int\frac{\sec{u}\cos^{2}(u)}{\sin^2(u)}du$$
$$=\frac{1}{9}\int\frac{\cos(u)}{\sin^2(u)}du.$$
$$=\frac{1}{9}\int\frac{1}{\sin^2(u)}d(\sin{u})$$
$$=\frac{1}{9}\int\sin^{-2}(u)d(\sin{u})$$
$$=\frac{1}{9}\left[-\sin^{-1}(u)\right]+C$$
Now we must return our expression in terms of $x$. That is, we wish to find the value of $\sin^{-1}(u)$ in terms of $x$. Note that this is not sine inverse - it is "sin(u) to the power of -1". Recall that:
$$ x = 3\tan{u} \implies \tan{u} = \frac{x}{3}.$$
Assume that $0 < u < \frac{\pi}{2}$ if you draw a right angled triangle you will notice that the opposite side is $x$ and the adjacent side is $3$. By Pythagoras's Theorem the hypotenuse is therefore $\sqrt{x^{2}+9}$. Therefore
$$\sin{u} = \frac{x}{\sqrt{x^{2}+9}}.$$
Thus, our answer is
$$\frac{1}{9}\left[-\frac{x}{\sqrt{x^{2}+9}}\right]^{-1}+C = -\frac{\sqrt{x^{2}+9}}{9x} +C .$$
