$\int \frac{1}{{(16+x^2)}^{3/2}}dx=$

Let $x=4\sin \theta ,$ Then $dx=4\cos \theta d\theta$

$\therefore \int \frac{dx}{{(16-x^2)}^{3/2}}=\int \frac{4\cos \theta }{{(16-16{\sin }^2\theta )}^{3/2}}d\theta$

$=\int \frac{4\cos \theta }{{\left(16(1-{\sin }^2\theta )\right)}^{3/2}} d\theta$

$=\int \frac{4\cos \theta }{64{\cos }^3\theta }d\theta$

$=\int \frac{1}{16}\frac{1}{{\cos }^2\theta }d\theta =\frac{1}{16}\int {\sec }^2\theta d\theta$

$=\frac{1}{16}\tan \theta +c$, where 'c' is constant of integration

$\sin ce x=4\sin \theta \Rightarrow \sin \theta =\frac{x}{4}\Rightarrow \tan \theta =\frac{x}{\sqrt{16-x^2}}+c$

$\therefore \int \frac{dx}{{(16-x^2)}^{\frac{3}{2}}}=\frac{x}{16\sqrt{16-x^2}}+c$