Never mind. I got the solution.
$\displaystyle \int{\frac{1}{(9 + x^2)^{\frac{3}{2}}}\,dx}$.
Make the substitution $\displaystyle x = 3\sinh{t}$ so that $\displaystyle dx = 3\cosh{t}\,dt$.
It's important to note that from our original substitution: $\displaystyle \sinh{t} = \frac{x}{3}$.
So the integral becomes
$\displaystyle \int{\frac{1}{[9 + (3\sinh{t})^2]^{\frac{3}{2}}}\,3\cosh{t}\,dt}$
$\displaystyle = \int{\frac{3\cosh{t}}{[9(1 + \sinh^2{t})]^{\frac{3}{2}}}\,dt}$
$\displaystyle = \int{\frac{3\cosh{t}}{27(\sqrt{\cosh^2{t}})^3}\,dt }$
$\displaystyle = \frac{1}{9}\int{\frac{\cosh{t}}{\cosh^3{t}}\,dt}$
$\displaystyle = \frac{1}{9}\int{\frac{1}{\cosh^2{t}}\,dt}$
$\displaystyle = \frac{1}{9}\tanh{t} + C$
$\displaystyle = \frac{\sinh{t}}{9\cosh{t}} + C$
$\displaystyle = \frac{\sinh{t}}{9\sqrt{1 + \sinh^2{t}}} + C$.
$\displaystyle = \frac{\frac{x}{3}}{9\sqrt{1 + \left(\frac{x}{3}\right)^2}} + C$
$\displaystyle = \frac{\frac{x}{3}}{9\sqrt{\frac{9 + x^2}{9}}} + C$
$\displaystyle = \frac{\frac{x}{3}}{\frac{9\sqrt{9 + x^2}}{3}} + C$
$\displaystyle = \frac{\frac{x}{3}}{3\sqrt{9 + x^2}} + C$
$\displaystyle = \frac{x}{9\sqrt{9 + x^2}} + C$.
Another alternative is to make the substitution $\displaystyle x = 3\tan{\theta}$.
The Hyperbolic sine.
Hyperbolic function - Wikipedia, the free encyclopedia
And like I said, substituting $\displaystyle x = 3\tan{\theta}$ should give you the same answer.