Answer
$$-\frac{x}{4 \sqrt{x^{2}-4}}+C$$
Work Step by Step
Given$$\int \frac{d x}{\left(x^{2}-4\right)^{\frac{3}{2}}} $$
Let, $x=2 \sec \theta, \quad d x=2 \sec \theta \tan \theta d \theta$
\begin{aligned}
\int \frac{d x}{\left(x^{2}-4\right)^{\frac{3}{2}}} &=\int \frac{2 \sec \theta \tan \theta d \theta}{\left.(2 \sec \theta)^{2}-4\right)^{\frac{3}{2}}} \\
&=\int \frac{2 \sec \theta \tan \theta d \theta}{\left(4 \sec ^{2} \theta-4\right)^{\frac{3}{2}}} \\
&=\int \frac{2 \sec \theta \tan \theta d \theta}{4^{\frac{3}{2}}\left(\sec ^{2} \theta-1\right)^{\frac{3}{2}}} \\
&=\int \frac{2 \sec \theta \tan \theta d \theta}{8 \tan ^{3} \theta} \\
&=\int \frac{2 \sec \theta d \theta}{8 \tan ^{2} \theta}\\
&=\frac{1}{4}\int \sin^{-2}\theta \cos \theta d\theta\\
&=\frac{-1}{4}\frac{1}{\sin \theta} +C\\
&=\frac{-1}{4}\frac{1}{\sqrt{1-\cos^2\theta}} +C\\
&=-\frac{x}{4 \sqrt{x^{2}-4}}+C
\end{aligned}
